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Unformatted text preview: COMMUTATIVE ALGEBRA
PETE L. CLARK Contents
Introduction
0.1. What is Commutative Algebra?
0.2. Why study Commutative Algebra?
0.3. Some themes of Commutative Algebra
1. Commutative rings
1.1. Fixing terminology
1.2. Adjoining elements
1.3. Ideals and quotient rings
1.4. Products of rings
1.5. A cheatsheet
2. Galois connections
2.1. The basics
2.2. Examples of Antitone Galois Connections
2.3. Examples of Isotone Galois Connections
3. Modules
3.1. Basic deﬁnitions
3.2. Finitely presented modules
3.3. Torsion and torsionfree modules
3.4. Tensor and Hom
3.5. Projective modules
3.6. Injective modules
3.7. Flat modules
3.8. Nakayama’s Lemma
3.9. Ordinal Filtrations and Applications
3.10. Tor and Ext
3.11. More on ﬂat modules
3.12. Faithful ﬂatness
4. First Properties of Ideals in a Commutative Ring
4.1. Introducing maximal and prime ideals
4.2. Radicals
4.3. Comaximal ideals
4.4. Local rings
5. Examples of rings
5.1. Rings of numbers
5.2. Rings of continuous functions
5.3. Rings of holomorphic functions
5.4. Polynomial rings
5.5. Semigroup algebras
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94 2 PETE L. CLARK 6. Swan’s Theorem
6.1. Introduction to (topological) vector bundles
6.2. Swan’s Theorem
6.3. Proof of Swan’s Theorem
6.4. Applications of Swan’s Theorem
7. Localization
7.1. Deﬁnition and ﬁrst properties
7.2. Pushing and pulling via a localization map
7.3. The ﬁbers of a morphism
7.4. Commutativity of localization and passage to a quotient
7.5. Localization at a prime ideal
7.6. Localization of modules
7.7. Local properties
7.8. Local characterization of ﬁnitely generated projective modules
8. Noetherian rings
8.1. Chain conditions on posets
8.2. Chain conditions on modules
8.3. Semisimple modules and rings
8.4. Normal Series
8.5. The KrullSchmidt Theorem
8.6. Some important terminology
8.7. Introducing Noetherian rings
8.8. The BassPapp Theorem
8.9. Artinian rings: structure theory
8.10. The Hilbert Basis Theorem
8.11. The Krull Intersection Theorem
8.12. Krull’s Principal Ideal Theorem
8.13. The Dimension Theorem
8.14. The ArtinTate Lemma
9. Boolean rings
9.1. Deﬁnition and ﬁrst properties of Boolean rings
9.2. Boolean Algebras
9.3. Ideal Theory in Boolean Rings
9.4. The Stone Representation Theorem
9.5. Boolean spaces
9.6. Stone Duality
9.7. Topology of Boolean Rings
10. Primary Decomposition
10.1. Some preliminaries on primary ideals
10.2. Primary decomposition, Lasker and Noether
10.3. Irredundant primary decompositions
10.4. Uniqueness properties of primary decomposition
10.5. Applications in dimension zero
10.6. Applications in dimension one
11. Aﬃne k algebras and the Nullstellensatz
11.1. Zariski’s Lemma
11.2. Hilbert’s Nullstellensatz
11.3. Other Nullstellens¨tze
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170 COMMUTATIVE ALGEBRA 12. Goldman domains and HilbertJacobson rings
12.1. Goldman domains
12.2. Hilbert rings
12.3. Jacobson Rings
12.4. HilbertJacobson Rings
13. The spectrum of a ring
13.1. The Zariski spectrum
13.2. Properties of the spectrum: quasicompactness
13.3. Properties of the spectrum: separation and specialization
13.4. Irreducible spaces
13.5. Noetherian spaces
13.6. The map on spectra induced by a homomorphism
13.7. Hochster’s Theorem
13.8. The support of a module
13.9. Rank functions revisited
14. Integrality in Ring Extensions
14.1. First properties of integral extensions
14.2. Integral closure of domains
14.3. Spectral properties of integral extensions
14.4. Integrally closed domains
14.5. The Noether Normalization Theorem
14.6. A Little Invariant Theory
14.7. Abstract algebraic number theory I: Galois extensions of integrally
closed domains
15. Factorization
15.1. Kaplansky’s Theorem (II)
15.2. Atomic domains, (ACCP)
15.3. ELdomains
15.4. GCDdomains
15.5. Polynomial rings over UFDs
15.6. Application: Eisenstein’s Criterion
15.7. Application: Determination of Spec R[t] for a PID R
15.8. Characterization of UFDs among Noetherian domains
15.9. Power series rings over UFDs
15.10. Nagata’s Criterion for UFDs
15.11. AuslanderBuchsbaum
16. Principal rings and B´zout domains
e
16.1. Principal ideal domains
16.2. A word on Euclidean functions
16.3. Some structure theory of principal rings
16.4. B´zout domains
e
17. Valuation rings
17.1. Basic theory
17.2. Ordered abelian groups
17.3. Connections with integral closure
17.4. Another proof of Zariski’s Lemma
17.5. Discrete valuation rings
18. Normalization theorems 3 170
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236 4 PETE L. CLARK 18.1. The First Normalization Theorem
18.2. The Second Normalization Theorem
18.3. The KrullAkizuki Theorem
19. Fractional ideals
19.1. Deﬁnition and ﬁrst properties
19.2. Principal Fractional Ideals
19.3. Invertible Fractional Ideals and the Picard Group
19.4. More on invertible ideals
19.5. Divisorial ideals
19.6. The Divisor Class Group
20. Dedekind domains
20.1. Characterization in terms of invertibility of ideals
20.2. Ideal factorization in Dedekind domains
20.3. Local characterization of Dedekind domains
20.4. Factorization into primes implies Dedekind
20.5. Generation of ideals in Dedekind domains
20.6. Finitely generated modules over a Dedekind domain
21. Pr¨fer domains
u
21.1. Characterizations of Pr¨fer Domains
u
21.2. Modules over a Pr¨fer domain
u
22. Structure of overrings
22.1. Introducing overrings
22.2. Overrings of Dedekind domains
22.3. Overrings of Pr¨fer Domains
u
22.4. Kaplansky’s Theorem (III)
22.5. Every commutative group is a Picard group
23. Krull domains
24. Connections with model theory
References 237
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261 Introduction
0.1. What is Commutative Algebra?
Commutative algebra is the study of commutative rings and attendant structures,
especially ideals and modules.
This is the only possible short answer I can think of, but it is not completely
satisfying. We might as well say that Hamlet, Prince of Denmark is about a ﬁctional royal family in late medieval Denmark and especially about the title (crown)
prince, whose father (i.e., the King) has recently died and whose father’s brother
has married his mother (i.e., the Queen). Informative, but not the whole story!
0.2. Why study Commutative Algebra?
What are the purely mathematical reasons for studying any subject of pure mathematics? I can think of two: COMMUTATIVE ALGEBRA 5 I. Commutative algebra is a necessary and/or useful prerequisite for the study
of other ﬁelds of mathematics in which we are interested.
II. We ﬁnd commutative algebra to be intrinsically interesting and we want to
learn more. Perhaps we even wish to discover new results in this area.
I think it is fair to say that most beginning students of commutative algebra can
relate to the ﬁrst reason: they need, or are told they need, to learn some commutative algebra for their study of other subjects. Indeed, commutative algebra has
come to occupy a remarkably central role in modern pure mathematics, perhaps
second only to category theory in its ubiquitousness, but in a diﬀerent way. (Basic)
category theory provides a common language and builds bridges between diﬀerent
areas of mathematics: it is something like a circulatory system. Commutative algebra provides core results that other results draw upon in a foundational way: it
is something like a skeleton.
The ﬁeld of mathematics which most of all draws upon commutative algebra for its
structural integrity is algebraic geometry, namely the study of geometric properties
of manifolds and singular spaces which arise as the loci of solutions to systems of
polynomial equations. In fact there is a hard lesson here: sometime in the 19th
century algebraic geometry split oﬀ from complex function theory and dierential
geometry as its own branch of mathematics and then burgeoned dramatically at the
turn of the (20th!) century and the years immediately thereafter. But by 1920 or so
the practitioners of the subject had found their way into territory in which “purely
geometric” reasoning led to serious errors. In particular they had been making
arguments about how algebraic varieties behave generically, but they lacked the
technology to even give a precise meaning to the term. Thus the subject ultimately
proved invertebrate and began to collapse under its own weight. Starting around
1930 there began a heroic shoring up process in which the foundations of the subject were recast with commutative algebraic methods at the core. In fact this was
done several times over, in diﬀerent ways, by Zariski, Weil, Serre and Grothendieck,
among others. For the last 60 years or so it has been impossible to study algebraic
geometry without knowing commutative algebra – a lot of commutative algebra.
(For instance, more than is contained in these notes!)
The other ﬁeld of mathematics which draws upon commutative algebra in an absolutely essential way is algebraic number theory. One sees this from the beginning
in that the Fundamental Theorem of Arithmetic is the assertion that the ring Z is
a unique factorization domain (UFD), a fundamental commutative algebraic concept. Moreover number theory was one of the historical sources of the subject.
Notably the concept of Dedekind domain came from Richard Dedekind’s numbertheoretic investigations. Knowledge of commutative algebra is not as absolutely
indispensable for number theory (at least, not at the beginning) as it is for algebraic geometry, but such knowledge brings a great clarifying eﬀect to the subject.
In fact the interplay among number theory, algebraic geometry and commutative
algebra ﬂows in all directions. What Grothendieck did in the 1960s (with important 6 PETE L. CLARK contributions from Chevalley, Serre and others) was to create a single ﬁeld of mathematics that encompassed commutative algebra and classical algebraic geometry –
and, as a byproduct, algebraic number theory: namely the theory of schemes. As
a result, most contemporary number theorists are also partly commutative algebraists and partly algebraic geometers: we call this more cosmopolitan take on the
subject arithmetic geometry.
But in fact there are plenty of other areas of mathematics that draw upon commutative algebra in slightly less dramatic but still important ways. To mention some
which will show up in later in these notes:
• Invariant theory.
• Diﬀerential topology.
• General topology.
• Order theory.
• Model theory.
The task of providing a commutative algebraic foundation for algebraic geometry alone – or even the single, seminal text of R. Hartshorne – is a daunting one.
Happily, this task has been completed by David Eisenbud (perhaps the leading
contemporary mathematician working at the interface of commutative algebra and
algebraic geometry) in his text [Eis]. This work is highly recommended. It is also
797 pages long, so contains enough material for 3 − 5 courses in the subject. It
would be folly to try to improve upon, or even successfully imitate, Eisenbud’s
work here, and I certainly have not tried.
I myself am an arithmetic geometer (which, as I tried to explain above, is a sort
of uppity kind of number theorist), so it is not surprising that these notes are
skewed more towards number theory than most introductory texts on commutative algebra. However for the most part a respectful distance is maintained: we
rarely discuss number theory per se but rather classes of rings that a number theorist would like: Dedekind domains, valuation rings, B´zout domains, and so forth.1
e
But just much as I have included some material of interest to number theorists
I have included material making connections to other branches of mathematics,
especially connections which are less traditionally made in commutative algebra
texts. In fact at several points I have digressed to discuss topics and theorems
which make connections to other areas of mathematics:
1If I may be so bold as to actually play one of my cards, for the past several years I have
been interested in developing a more commutative algebraic take on classical algebraic number
theory, which would have the pleasant consequence of developing the function ﬁeld case not just
in parallel with the number ﬁeld case but as a second specialization of a more general theory. The
sucess of this task is nowhere guaranteed and at any rate is not very near at hand. This text is,
in part, a very preliminary step in this direction. COMMUTATIVE ALGEBRA •
•
•
•
•
• §X.X
§X.X
§X.X
§X.X
§X.X
§X.X on
on
on
on
on
on 7 Galois connections.
rings of continuous functions.
vector bundles and Swan’s Theorem.
Boolean rings, Boolean spaces and Stone Duality.
the topology of prime spectra, including Hochster’s Theorem.
invariant theory, including the ShephardToddChevalleyTheorem. As a whole, general topological issues arise surprisingly often in these notes. This
was not by design, but seems to point to a kinship between these two ﬁelds, aspects
of which are cetainly known to the experts.
On the other hand, I do ﬁnd commutative algebra to be of interest unto itself,
and I have tried to craft a sustained narrative in which various basic questions get
explored in increasing depth and certain structures recur. Like any bad writer, I
am not conﬁdent that I have made these themes apparent in the text itself, so in
the following section I will explicitly point some of them out to you.
0.3. Some themes of Commutative Algebra.
1. Commutative rings
1.1. Fixing terminology.
We are interested in studying properties of rings, generally commutative rings
and invariably rings with unity. So let us begin by clarifying this terminology.
By a general algebra R, we mean a triple (R, +, ·) where R is a set endowed
with a binary operation + : R × R → R – called addition – and a binary operation
· : R × R → R – called multiplication – satisfying the following:
(CG) (R, +) is a commutative group,
(D) For all a, b, c ∈ R, (a + b) · c = a · c + b · c, a · (b + c) = a · b + a · c.
For at least ﬁfty years, there has been agreement that in order for an algebra
to be a ring, it must satisfy the additional axiom of associativity of multiplication:
(AM) For all a, b, c ∈ R, a · (b · c) = (a · b) · c.
A general algebra which satisﬁes (AM) will be called simply an algebra. A similar
convention that is prevalent in the literature is the use of the term nonassociative
algebra to mean what we have called a general algebra: i.e., a not necessarily
associative algebra.
A ring R is said to be with unity if there exists a multiplicative identity, i.e.,
an element e of R such that for all a ∈ R we have e · a = a · e = a. If e and e′
are two such elements, then e = e · e′ = e′ . In other words, if a unity exists, it is
unique, and we will denote it by 1. 8 PETE L. CLARK A ring R is commutative if for all x, y ∈ R, x · y = y · x.
With very few exceptions – which will be made clear! – we will be working always in the category of commutative rings with unity. In a sense which will be
made precise shortly, this means that the identity 1 is regarded as a part of the
structure of a ring, and must therefore be preserved by all homomorphisms of rings.
Probably it would be more natural to study the class of possibly noncommutative
rings with unity, since, as we will see, many of the fundamental constructions of
rings give rise, in general, to noncommutative rings. But if the restriction to
commutative rings (with unity!) is an artiﬁce, it is a very useful one, since two
of the most fundamental notions in the theory, that of ideal and module, become
signiﬁcantly diﬀerent and more complicated in the noncommutative case. It is
nevertheless true that many individual results have simple analogues in the noncommutative case. But it does not seem necessary to carry along the extra generality of noncommutative rings; rather, when one is interested in the noncommutative
case, one can simply remark “Proposition X.Y holds for (left) Rmodules over a
noncommutative ring R.”
Notation: Generally we shall abbreviate x · y to xy . Moreover, we usually do
not use diﬀerent symbols to denote the operations of addition and multiplication
in diﬀerent rings: it will be seen that this leads to simplicity rather than confusion.
Group of units: Let R be a ring with unity. An element x ∈ R is said to be a
unit if there exists an element y such that xy = yx = 1.
Exercise X.X:
a) Show that if x is a unit, the element y with xy = yx = 1 is unique, denoted x−1 .
b) Show that if x is a unit, so is x−1 .
c) Show that, for all x, y ∈ R, xy is a unit ⇐⇒ x and y are both units. d) Deduce
that the units form a group, denoted R× , under multiplication.
e) Give an example of an element x in a ring R for which there exists y such that
xy = 1 but there does not exist an element z such that zx = 1.
Example (Zero ring): Our rings come with two distinguished elements, the additive identity 0 and the multiplicative identity 1. Suppose that 0 = 1. Then for
x ∈ R, x = 1 · x = 0 · x, whereas in any rin g 0 · x = (0 + 0) · x = 0 · x + 0 · x, so
0 · x = 0. In other words, if 0 = 1, then this is the only element in the ring. It is
clear that for any one element set R = {0}, 0 + 0 = 0 · 0 = 0 endows R with the
structure of a ring. We call this ring the zero ring.
The zero ring exhibits some strange behavior, such that it must be explicitly excluded in many results. For instance, the zero element is a unit in the zero ring,
which is obviously not the case in any nonzero ring. A nonzero ring in which every
nonzero element is a unit is called a division ring. A commutative division ring
is called a ﬁeld.
Let R and S be rings (with unity). A homomorphism f : R → S is a map COMMUTATIVE ALGEBRA 9 of sets which satisﬁes
(HOM1) For all x, y ∈ R, f (x + y ) = f (x) + f (y ).
(HOM2) For all x, y ∈ R, f (xy ) = f (x)f (y ).
(HOM3) f (1) = 1.
Note that (HOM1) implies f (0) = f (0 + 0) = f (0) + f (0), so f (0) = 0. Thus
we do not need to explcitly include f (0) = 0 in the deﬁnition of a group homomorphism. For the multiplicative identity however, this argument only shows that if
f (1) is a unit, then f (1) = 1. Therefore, if we did not require (HOM3), then for
instance the map f : R → R, f (x) = 0 for all x, would be a homomorphism, and
we do not want this.
Exercise X.X: Suppose R and S are rings, and let f : R → S be a map satisfying
(HOM1) and (HOM2). Show that f is a homomorphism of rings (i.e., satisﬁes also
f (1) = 1) iﬀ f (1) ∈ S × .
A homomorphism f : R → S is an isomorphism if there exists a homomorphism
g : S → R such that: for all x ∈ R, g (f (x)) = x; and for all y ∈ S , f (g (y )) = y .
Exercise X.X: Let f : R → S be a homomorphism of rings. Show that TFAE:
(i) f is a bijection.
(ii) f is an isomorphism.
Remark: In many algebra texts, an isomorphism of rings (or groups, etc.) is deﬁned to be a bijective homomorphism, but this gives the wrong idea of what an
isomorphism should be in other mathematical contexts (e.g. for topological spaces).
Rather, having deﬁned the notion of a morphism of any kind, one deﬁnes isomorphism in the way we have above.
Exercise X.X.X: a) Suppose R and S are both rings on a set containing exactly
one element. Show that there is a unique ring isomorphism from R to S . (This is
a triviality, but explains why are we able to speak of the zero ring, rather than
simply the zero ring associated to one element set. We will therefore denote the
zero ring just by 0.)
b) Show that any ring R admits a unique homomorphism to the zero ring. One
says that the zero ring is the ﬁnal object in the category of rings.
Exercise X.X.X: Show that for any ring S there exists a unique homomorphism
from the ring Z of integers to S . (One says that Z is the initial object in the
category of rings.)
A subring R of a ring S is a subset R of S such that
(SR1) 1 ∈ R.
(SR2) For all r, s ∈ R, r + s R, r − s ∈ R, and rs ∈ R. 10 PETE L. CLARK Here (SR2) expresses that the subset R is an algebra under the operations of addition and multiplication deﬁned on S . Working, as we are, with rings with unity, we
have to be a bit more careful: in the presence of (SR2) but not (SR1) it is possible
that R either does not have a multiplicative identity or, more subtly, that it has a
multiplicative identity which is not the element 1 ∈ S .
An example of the ﬁrst phenomenon is S = Z, R = 2Z. An example of the
second is S = Z, R = 0. A more interesting example is S = Z × Z – i.e., the set
of all ordered pairs (x, y ), x, y ∈ Z with (x1 , y1 ) + (x2 , y2 ) = (x1 + x2 , y1 + y2 ),
(x1 , y1 ) · (x2 , y2 ) = (x1 x2 , y1 y2 ) – and R = {(0, y )  y ∈ Z}. Then with the induced
addition and multiplication from S , R is isomorphic to the ring Z and the element
(0, 1) serves as a multiplicative identity on R which is diﬀerent from the (always
unique) multiplicative identity 1S = (1, 1), so according to our conventions R is not
a subring of S .
Notice that if R is a subring of S , the inclusion map R → S is an injective homomorphism of rings. Conversely, if ι : R → S is an injective ring homomorphism,
then R ∼ ι(R) and ι(R) is a subring of S , so essentially we may use ι to view R
=
as a subring of S . The only proviso here is that this certainly depends on ι: in
general there may be other injective homomorphisms ι : R → S which realize R as
a diﬀerent subset of S , hence a diﬀerent subring.
1.2. Adjoining elements.
Let ι : R → S be an injective ring homomorphism. As above, let us use ι to
view R as a subring of S ; we also say that S is an extension ring of R and write
S/R for this (note: this has nothing to do with cosets or quotients!) We wish now
to consider rings T such that R ⊂ T ⊂ S ; such a ring T might be called a subextension of S/R or an intermediate ring.
Let X = {xi } be a subset of S . Then the partially ordered set of all subrings
of T containing R and X contains a bottom element, given (as usual!) by taking
the intersection of all of its elements. (This partially ordered set is nonempty, since
S is in it.) We call this the ring obtained by adjoining the elements of X to R. In
the commutative case, we denote this ring by R[{xi }], for reasons that will become
more clear when we discuss polynomial rings in §X.X .
√
Example: Take R = Z, S = C. Then Z[i] = Z[ −1] is the smallest subring of
√
C containing (Z and) −1.
Example: Take R = Z, S = Q, let P be any set of prime numbers, and put
1
1
X = { p }p∈P . Then there is a subring ZP := Z[{ p }p∈P ] of Q.
Exercise: Let P , Q be two sets of prime numbers. Show TFAE:
(i) ZP ∼ ZQ .
=
(ii) ZP = ZQ .
(iii) P = Q.
Exercise: Show that every subring of Q is of the form ZP for some P . COMMUTATIVE ALGEBRA 11 Note well that the adjunction process R → R[X ] is deﬁned only relative to some
extension ring S of R, although the notation hides this. In fact, one of the recurrent
themes of the subject is the expression of the adjunction process in a way which
√
depends only on R itself. In the ﬁrst example, this is √
achieved by identifying −1
with its minimal polynomial t2 + 1 and replacing Z[ −1] with the quotient ring
Z[t]/(t2 + 1). The second example will eventually be turned around: we will be
able to give an independent deﬁnition of ZP as a certain “ring of fractions” formed
from Z and then Q will be the ring of fractions obtained by taking P to be the set
of all prime numbers.
Nevertheless, the existence of such turnabouts should not cause us to forget that
adjunction is relative to √ extension; indeed forgetting this can lead to serious
an
trouble. For instance, if 3 2 is the unique real cube root of 2 and ζ3 is a primitive
√
cube √ of unity, then the three complex numbers with cube 2 are z1 = 3 2,
root
√2
z2 = 3 2ζ3 and z3 = 3 2ζ3 . Each of the rings Q[z1 ], Q[z2 ], Q[z3 ] is isomorphic to
the ring Q[t]/(t3 − 2), so all three are isomorphic to each other. But they are not
the same ring: on the one hand Q[z1 ] is contained in R and the other two are not.
√
More seriously Q[z1 , z2 , z3 ] = Q[ 3 2, ζ3 ], which strictly contains any one of Q[z1 ],
Q[z2 ] and Q[z3 ].
1.3. Ideals and quotient rings.
Let f : R → S be a homomorphism of rings, and put
I = f −1 (0) = {x ∈ R  f (x) = 0}.
Then, since f is in particular a homomorphism of commutative groups (R, +) →
(S, +), I is a subgroup of (R, +). Moreover, it enjoys both of the following properties:
(LI) For all i ∈ I and y ∈ R, iy ∈ I .
(RI) For all j ∈ I and x ∈ R, xj ∈ I .
Indeed,
f (xj ) = f (x)f (j ) = f (x) · 0 = 0 = 0 · f (y ) = f (i)f (y ) = f (iy ).
In general, let R be a ring. An ideal is a subset I ⊂ R which is a subgroup of
(R, +) (in particular, 0 ∈ I ) and which satisﬁes (LI) and (RI).
Theorem 1. Let R be a ring, and let I be a subgroup of (R, +). TFAE:
(i) I is an ideal of R.
(ii) There exists a ring structure on the quotient group R/I making the additive
homomorphism R → R/I into a homomorphism of rings.
When these conditions hold, the ring structure on R/I in (ii) is unique, and R/I
is called the quotient of R by the ideal I .
Proof. Consider the group homomorphism q : R → R/I . If we wish R/I to be a
ring in such a way so that q is a ring homomorphism, we need
(x + I )(y + I ) = q (x)q (y ) = q (xy ) = (xy + I ). 12 PETE L. CLARK This shows that there is only one possible ring structure, and the only question is
whether it is welldeﬁned. For this we need that for all i, j ∈ I , (x + i)(y + j ) − xy =
xj + iy + ij ∈ I . Evidently this holds for all x, y, i, j iﬀ (LI) and (RI) both hold.
Remark: If R is commutative, then of course there is no diﬀerence between (LI) and
(RI). For a noncommutative ring R, an additive subgroup I satisfying condition
(LI) but not necessarily (RI) (resp. (RI) but not necessarily (LI)) is called a left
ideal (resp. a right ideal). Often one says twosided ideal to emphasize that
(LI) and (RI) both hold. Much of the additional complexity of the noncommutative
theory comes from the need to distinguish between left, right and twosided ideals.
We do not wish to discuss such complexities here, so henceforth in this section
we assume (except in exercises, when indicated) that our rings are commutative.
Example: In R = Z, for any integer n, consider the subset (n) = nZ = {nx  x ∈ Z}
of all multiples of n. This is easily seen to be an ideal.2 The quotient Z/nZ is the
ring of integers modulo n.
An ideal I R is called proper. Exercise X.X: Let R be a ring and I an ideal of R. Show that TFAE:
(i) I ∩ R× ̸= ∅.
(ii) I = R.
Exercise X.X: a) Let R be a commutative ring. Show that R is a ﬁeld iﬀ R has
exactly two ideals, 0 and R.
b) Let R be a not necessarily commutative ring. Show that TFAE: (i) The only
onesided ideals of R are 0 and R. (ii) R is a division ring.
c) For a ﬁeld k and an integer n > 1, show that the matrix ring Mn (k ) has no
twosided ideals but is not a division ring.
Exercise X.X: In contemporary undergraduate algebra texts, there is a trend to
deﬁne the ﬁnite ring Z/nZ in a diﬀerent and apparently simpler way: put Zn =
{0, 1, . . . , n − 1}. For any integer x, there is a unique integer k such that x − kn ∈ Zn .
Deﬁne a function mod n : Z → Zn by mod n(x) := x − kn. We then deﬁne +
and · on Zn by x + y := mod n(x + y ), xy = mod n(xy ). Thus we have avoided
any mention of ideals, equivalence classes, quotients, etc. Is this actually simpler?
(Hint: how do we know that Zn satisﬁes the ring axioms?)
For any commutative ring R and any element y ∈ R, the subset (y ) = yR =
{xy  x ∈ R} is an ideal of R. Such ideals are called principal. A principal ideal
ring is a commutative ring in which each ideal is principal.
Exercise X.X: a) The intersection of an arbitrary family of (left, right or twosided)
ideals in a possibly noncommutative ring is a (left, right or twosided) ideal.
∩
b) Let {Ii } be a set of ideals in the commutative ring R. Show that ∩ i has the
iI
following property: for any ideal J of R such that J ⊂ Ii for all i, J ⊂ i I .
2If this is not known and/or obvious to the reader, these notes will probably be too brisk. COMMUTATIVE ALGEBRA 13 Let R be a ring and S a subset of R. There is then a smallest ideal of R con∩
taining S , namely Ii , where Ii are all the ideals of R containing S . We call this
the ideal generated by S . This is a “topdown” description; as usual, there is a
complementary “bottomup” description which is not quite as clean but often more
useful. Namely, put
∑
⟨S ⟩ := {
ri si  ri ∈ R, si ∈ S }
i.e., the set of all ﬁnite sums of an element of R times an element of S .
Proposition 2. For a subset S of a commutative ring R, ⟨S ⟩ is an ideal, the
intersection of all ideals containing S .
Exercise X.X: Prove Proposition X.X.
Exercise X.X: Let R be a ring.
a) For ideals I and J of R, deﬁne I + J = {i + j  i ∈ I, j ∈ J }. Show that
I + J = ⟨I ∪ J ⟩ is the smallest ideal containing both I and J .
b) Extend part a) to any ﬁnite number of ideals I1 , . . . , In .
c) Suppose {Ii } is an arbitrary set of ideals of I . Give an explicit description of the
ideal ⟨Ii ⟩.
Remark: The preceding considerations show that the collection of all ideals of
a commutative ring R, partially ordered by inclusion, form a complete lattice.
If I is an ideal in the ring R, then there is a correspondence between ideals J
of R containing I and ideals of the quotient ring R/I , exactly as in the case of a
normal subgroup of a group:
Theorem 3. (Correspondence theorem) Let I be an ideal of a ring R, and denote
the quotient map R → R/I by q . Let I (R) be the lattice of ideals of R, II (R) be
the sublattice of ideals containing I and I (R/I ) the lattice of ideals of the quotient
ring R/I . Deﬁne maps
Φ : I (R) → I (R/I ), J → (I + J )/I,
Ψ : I (R/I ) → I (R), J → q −1 (J ).
Then Ψ ◦ Φ(J ) = I + J and Φ ◦ Ψ(J ) = J . In particular Ψ induces an isomorphism
of lattices from I (R/I ) to II (R).
Proof. For all the abstraction, the proof is almost trivial. For J ∈ I (R), we check
that Ψ(Φ(J )) = Ψ(J + I (mod I )) = {x ∈ R  x + I ∈ J + I } = J + I ∈ II (R).
Similarly, for J ∈ I (R/I ), we have Φ(Ψ(J )) = J .
Remark: In fancier language, the pair (Φ, Ψ) give an isotone Galois connection
between the partially ordered sets I (R) and I (R/I ). The associated closure operator Φ ◦ Ψ on I (R/I ) is the identity, whereas the closure operator Ψ ◦ Φ on I (R)
carries each ideal J to the smallest ideal containing both J and I .
The correspondence theorem will be a constant companion in our study of rings
and ideals. As is common, we will often use the map Ψ to identify the sets I (R/I )
and II (R). 14 PETE L. CLARK Exercise X.X: Let I be an ideal of R and {Ji } be a set of ideals of R. Show
that Φ preserves suprema and Ψ preserves inﬁma:
Φ(⟨Ji ⟩) = ⟨Φ(Ji )⟩
and ∩
∩
Ψ( Ji ) =
Ψ(Ji ). 1.4. Products of rings.
Let R1 and R2 be rings. The Cartesian product R1 × R2 has the structure of
a ring with “componentwise” addition and multiplication:
(r1 , r2 ) + (s1 , s2 ) := (r1 + s1 , r2 + s2 ).
(r1 , r2 ) · (s1 , s2 ) := (r1 s1 , r2 s2 ).
Exercise X.X.X:
a) Show that R1 × R2 is commutative iﬀ both R1 and R2 are commutative.
b) R1 × R2 has an identity iﬀ both R1 and R2 do, in which case e := (e1 , e2 ) is the
identity of R1 × R2 .
As for any Cartesian product, R1 × R2 comes equipped with its projections
π1 : R1 × R2 → R1 ,  (r1 , r2 ) → r1
π2 : R1 × R2 → R2 ,  (r1 , r2 ) → r2 .
The Cartesian product X1 × X2 of sets X1 and X2 satisﬁes the following universal
property: for any set Z and any maps f1 : Z → X1 , f2 : Z → X2 , there exists a
unique map f : Z → X1 × X2 such that f1 = π1 ◦ f , f2 = π2 ◦ f . The Cartesian
product R1 × R2 satisﬁes the analogous universal property in the category of rings:
Exercise X.X: For rings R1 , R2 , S and ring homomorphisms fi : S → Ri , there
exists a unique homomorphism of rings f : S → R1 × R2 such that fi = πi ◦ f .
So the Cartesian product of R1 and R2 is also the product in the categorical sense.
As with sets, we can equally well take the Cartesian product over an arbitrary
indexed family of rings: if {Ri }i∈I is a family of rings, their Cartesian product
∏
i∈I Ri becomes a ring under coordinatewise addition and multiplication, and satisﬁes the universal property of the product. Details are left to the reader.
It is natural to ask whether the category of rings has a direct sum as well. In other
words, given rings R1 and R2 we are looking for a ring R together with ring homomorphisms ιi : Ri → R such that for any ring S and homomorphisms fi : Ri → S ,
there exists a unique homomorphism f : R → S such that fi = f ◦ ιi . We recall
that in the category of abelian groups, the Cartesian product group G1 × G2 also
the categorical direct sum, with ι1 : g → (g, 0) and ι2 : g → (0, g ). Since each ring
has in particular the structure of an abelian group, it is natural to wonder whether
the same might hold true for rings. However, the map ι1 : R1 → R1 × R2 does not
preserve the multiplicative identity (unless R2 = 0), so is not a homomorphism of
rings when identities are present. Moreover, even in the category of algebras, in
order to satisfy the universal property on the underlying additive subgroups, the COMMUTATIVE ALGEBRA 15 homomorphism f is uniquely determined to be (r1 , r2 ) → f1 (r1 ) + f2 (r2 ), and it is
easily checked that this generally does not preserve the product.
We will see later that the category of rings does have direct sums in the categorical sense: the categorical direct sum of R1 and R2 is given by the tensor product
R1 ⊗Z R2 .
Now returning to the case of commutative rings, let us consider the ideal structure of the product R = R1 × R2 . It is easy to see that if I1 is an ideal of R1 , then
I1 × {0} = {(i, 0)  i ∈ I } is an ideal of the product; moreover the quotient R/I1
is isomorphic to R1 /I1 × R2 . Similarly, if I2 is an ideal, {0} × I2 is an ideal of R2 .
Finally, if I1 is an ideal of R1 and I2 is an ideal of R2 , then
I1 × I2 := {(i1 , i2 ) i1 ∈ I1 , i2 ∈ I2 }
is an ideal of R. In fact we have already found all the ideals of the product ring:
Proposition 4. Let R1 and R2 be commutative rings, and let I be an ideal of
R := R1 × R2 . Put
I1 := {r1 ∈ R1  ∃r2 ∈ R2  (r1 , r2 ) ∈ I },
I2 := {r2 ∈ R2  ∃r1 ∈ R1  (r1 , r2 ) ∈ I }.
Then I = I1 × I2 = {(i1 , i2 )  i1 ∈ I1 , i2 ∈ I2 }.
Proof. Observe ﬁrst that I1 × {0} and {0} × I2 are ideals of R contained in I .
Indeed, if i1 ∈ I1 , then (i1 , r2 ) ∈ I for some r2 and then (i1 , 0) = (i1 , r2 ) · (1, 0),
and similarly for I2 . Therefore
I1 × I2 = (I1 × {0}) + ({0} × I2 ) ⊂ I.
Conversely, if (x, y ) ∈ I , then
(x, y ) = (x, 0)(1, 0) + (0, y )(0, 1) ∈ I1 × I2 .
Remark: Another way to express the result is that, corresponding to a decomposition R = R1 × R2 , we get a decomposition I (R) = I (R1 ) × I (R2 ).
Let us call a commutative ring R disconnected if there exists nonzero rings R1 ,
R2 such that R ∼ R1 × R2 , and connected otherwise.3 If R is disconnected,
=
then choosing such an isomorphism φ, we may put I1 = φ−1 (R1 × {0}) and
I2 = φ−1 ({0} × R2 ). Evidently I1 and I2 are ideals of R such that I1 ∩ I2 = {0} and
I1 × I2 = R. Conversely, if in a ring R we can ﬁnd a pair of ideals I1 , I2 with these
properties then it will follow from the Chinese Remainder Theorem (Theorem 126)
that the natural map Φ : R → R/I2 × R/I1 , r → (r + I2 , r + I1 ) is an isomorphism.
Now Φ restricted to I1 induces an isomorphism of groups onto R/I2 (and similarly
with the roles of I1 and I2 reversed). We therefore have a distinguished element
of I1 , e1 := Φ−1 (1). This element e1 is an identity for the multiplication on R restricted to I1 ; in particular e2 = e1 ; such an element is called an idempotent. In
1
3We will see later that there is a topological space Spec R associated to every ring, such that
Spec R is disconnected in the usual topological sense iﬀ R can be written as a nontrivial product
of rings 16 PETE L. CLARK any ring the elements 0 and 1 are idempotents, called trivial; since e1 = Φ−1 (1, 0) –
and not the preimage of (0, 0) or of (1, 1) – e1 is a nontrivial idempotent. Thus
a nontrivial decomposition of a ring implies the presence of nontrivial idempotents.
The converse is also true:
Proposition 5. Suppose R is a ring and e is a nontrivial idempotent element of
R: e2 = e but e ̸= 0, 1. Put I1 = Re and I2 = R(1 − e). Then I1 and I2 are ideals
of R such that I1 ∩ I2 = 0 and R = I1 + I2 , and therefore R ∼ R/I1 × R/I2 is a
=
nontrivial decomposition of R.
Exercise X.X: Prove Proposition X.X.
Exercise X.X: Generalize the preceding discussion to decompositions into a ﬁnite
number of factors: R = R1 × · · · × Rn .
1.5. A cheatsheet.
Let R be a commutative ring. Here are some terms that we will analyze in loving detail later, but would like to be able to mention in passing whenever necessary.
R is an integral domain if xy = 0 =⇒ x = 0 or y = 0.
An ideal p of R is prime if the quotient ring R/p is an integral domain. Equivalently, p is an ideal such that xy ∈ p =⇒ x ∈ p or y ∈ p.
An ideal m of R is maximal if it is proper – i.e., not R itself – and not strictly
contained in any larger proper ideal. Equivalently, m is an ideal such that the quotient ring R/m is a ﬁeld.
R is Noetherian if it satisﬁes any of the following equivalent conditions:4
(i) For any nonempty set S of ideals of R, there exists I ∈ S which is not properly
contained in any J ∈ S .
(ii) There is no inﬁnite sequence of ideals I1 I2 . . . in R.
(iii) Every ideal of R is ﬁnitely generated.
R is Artinian (or sometimes, an Artin ring) if the partially ordered set of ideals
of R satisﬁes the descending chain condition: there is no inﬁnite sequence of ideals
I1 I2 . . ..
If I and J are ideals of a ring R, we deﬁne the colon ideal5
(I : J ) = {x ∈ R  xJ ⊂ I }.
(I : J ) is also an ideal.
4See Theorem 208 for a proof of their equivalence.
5The terminology is unpleasant and is generally avoided as much as possible. One should think of (I : J ) as being something like the “ideal quotient” I/J (which of course has no formal
meaning). Its uses will gradually become clear. COMMUTATIVE ALGEBRA 17 2. Galois connections
2.1. The basics.
Let (X, ≤) and (Y, ≤) be partially ordered sets. A map f : X → Y is isotone
(or orderpreserving) if for all x1 , x2 ∈ X , x1 ≤ x2 =⇒ f (x1 ) ≤ f (x2 ). A map
f : X → Y is antitone (or orderreversing) if for all x1 , x2 ∈ X, x1 ≤ x2 =⇒
f (x1 ) ≥ f (x2 ).
Let (X, ≤) and (Y, ≤) be partially ordered sets. An antitone Galois connection between X and Y is a pair of maps Φ : X → Y and Ψ : Y → X such that:
(GC1) Φ and Ψ are both antitone maps, and
(GC2) For all x ∈ X and all y ∈ Y , x ≤ Ψ(y ) ⇐⇒ Φ(x) ≥ y .
Remark 1.1: There is a pleasant symmetry between X and Y in the deﬁnition.
That is, if (Φ, Ψ) is a Galois connection between X and Y , then (Ψ, Φ) is a Galois
connection between Y and X . Here is a fancy rephrasing: every Galois connection
(Φ, Ψ) has a dual Galois connection (Ψ, Φ).
Here is a good exercise for the reader trying to wrap her mind around Galois
connections for the ﬁrst time. I can (virtually) guarantee that somewhere in your
past lies a selfdual Galois connection, i.e., with X = Y and Φ = Ψ. What is it?
Many examples will be given later on. First we want to establish the fundamental
properties of any Galois connection.
Deﬁnition: If (X, ≤) is a partially ordered set, then a mapping f : X → X is
called a closure operator if it satisﬁes all of the following properties:
(C1) For all x ∈ X , x ≤ f (x).
(C2) For all x1 , x2 ∈ X , x1 ≤ x2 =⇒ f (x1 ) ≤ f (x2 ).
(C3) For all x ∈ X , f (f (x)) = f (x).
Proposition 6. The mapping Ψ◦Φ is a closure operator on (X, ≤) and the mapping
Φ ◦ Ψ is a closure operator on (Y, ≤).
Proof. Symmetry considerations (c.f. Remark 1.1) show that it is enough to consider the mapping x → Ψ(Φ(x)) on X .
If x1 ≤ x2 , then since both Φ and Ψ are antitone, we have Φ(x1 ) ≥ Φ(x2 ) and
thus Ψ(Φ(x1 )) ≤ Ψ(Φ(x1 )): (C2).6
For x ∈ X , Φ(x) ≥ Φ(x), and by (GC2) this implies x ≤ Ψ(Φ(x)): (C1).
Finally, for x ∈ X , applying (C1) to the element Ψ(Φ(x)) of X gives
Ψ(Φ(x)) ≤ Ψ(Φ(Ψ(Φ(x)))).
Conversely, we have
Ψ(Φ(x)) ≤ Ψ(Φ(x)),
so by (GC2)
Φ(Ψ(Φ(x)) ≥ Φ(x),
6In other words, the composition of two antitone maps is isotone. 18 PETE L. CLARK and applying the orderreversing map Ψ gives
Ψ(Φ(Ψ(Φ(x)))) ≤ Ψ(Φ(x)).
Thus
Ψ(Φ(x)) = Ψ(Φ(Ψ(Φ(x))).
Corollary 7. The operators (Φ, Ψ) in a Galois Connection satisfy the following
tridempotence properties:
a) For all x ∈ X , ΦΨΦx = Φx.
b) For all y ∈ X , ΨΦΨy = Ψy .
Exercise: Prove Corollary 7.
Proposition 8. Let (Φ, Ψ) be a Galois connection between partially ordered sets
X and Y . Let X = Ψ(Φ(X )) and Y = Ψ(Φ(Y )).
a) X and Y are precisely the subsets of closed elements of X and Y respectively.
b) We have Φ(X ) ⊂ Y and Ψ(Y ) ⊂ X .
c) Φ : X → Y and Ψ : Y → X are mutually inverse bijections.
Proof. a) If x = Ψ(Φ(x)) then x ∈ X . Conversely, if x ∈ X , then x = Ψ(Φ(x′ )) for
some x′ ∈ X , so Ψ(Φ(x))) = Ψ(Φ(Ψ(Φ(x′ )))) = Ψ(Φ(x′ )) = x, so X is closed.
b) This is just a reformulation of Corollary 7.
c) If x ∈ X and y ∈ Y , then Ψ(Φ(x)) = x and Ψ(Φ(y )) = y .
We speak of the mutually inverse antitone bijections Φ : X → Y and Ψ : Y → X
as the Galois correspondence induced by the Galois connection (Φ, Ψ).
In applications, the content usually lies in determining the subsets X and Y .
Having established the basic results, we will now generally abbreviate the closure
operators Ψ ◦ Φ and Φ ◦ Ψ to x → x and y → y .
2.2. Examples of Antitone Galois Connections.
If X is a set, then the power set 2X (of all subsets of X ) is partially ordered
via containment: for Y1 , Y2 ⊂ X , put Y1 ≤ Y2 if Y1 ⊂ Y2 . If A ⊂ 2X is any subset,
then it gets an induced ordering by containment. In all of our examples of Galois
connections below, the partially ordered sets X and Y are both of this form.
Example 1 (Galois connection of a Gset): Let G be a group acting on a set X .
Let X = 2X be the collection of all subsets of X and G be the collection of all
subgroups of G (each made into a partially ordered set by inclusion, as above). We
deﬁne Φ : X → G by
Φ : A ⊂ X → {g ∈ G  ∀x ∈ A, gx = x}
and Ψ : G → X by
Ψ : H ⊂ G → X H = {x ∈ X  ∀g ∈ H, gx = x}.
Let us check that (Φ, Ψ) is a Galois connection. It is easy to see (and probably
familiar) that Φ and Ψ are antitone: a smaller set has a larger subgroup acting
trivially on it, and a smaller subgroup has a larger set of ﬁxed points. Now let Y COMMUTATIVE ALGEBRA 19 be a subset of X and H a subgroup of G, and let us unpack the condition that
A ≤ Ψ(H ) ⇐⇒ Φ(A) ≥ H . In this case, A ≤ Ψ(H ) says that for every y ∈ A
and every h ∈ H , hy = y , whereas Φ(A) ≥ H says that for every h ∈ H and every
y ∈ A, hy = y . These two statements are plainly equivalent!
Subexample 1.1: Let L/K be a ﬁeld extension, X = L and G = Aut(L/K ). Then
it is not hard to show that every ﬁnite subgroup of G is closed. On the other hand,
every closed subset of X is a subextension F of L/K , but it is not necessarily the
case that every subextension is closed. If L/K is ﬁnite, then so is G and thus every
subgroup of G is closed. On the other side, much of the content of classical Galois
theory resides in the fact that every subextension of L/K is closed iﬀ L/K is normal
and separable. In the case of L/K inﬁnite algebraic, it is again the case that every
subextension F of L/K is closed iﬀ L/K is normal and separable. However, it is
no longer the case that every subgroup of Aut(L/K ) is closed. Rather, the closed
subgroups of Aut(L/K ) form a local base at the identity for a proﬁnite topology
on Aut(L/K ), the Krull topology.
Example 2 (Galois connection on a set of functions) Let X and R be sets and
S a subset of R. Let F ⊂ Map(X, R) be a set of functions f : X → R. We deﬁne
a Galois connection (Φ, Ψ) between 2X and 2F . For V ⊂ X , we deﬁne
Φ(V ) = {f ∈ F  f (V ) ⊂ S }.
For I ⊂ F , we deﬁne
Ψ(I ) = {x ∈ X ∀f ∈ I, f (x) ∈ S }.
Again it is easy to see that Φ and Ψ are antitone. For V ⊂ X and I ⊂ F , the
condition V ⊂ Ψ(I ) means that for all x ∈ V and all f ∈ I , f (x) ∈ S , whereas the
condition I ⊂ Φ(V ) means that for all f ∈ I and all x ∈ V , f (x) ∈ S . These two
statements are plainly equivalent!
Suppose that R is an abelian group and S is a subgroup of R. Then Map(X, R) is
an abelian group under pointwise addition. Further suppose that F is a subgroup
of Map(X, R). Then for each V ⊂ X , Φ(V ) is the subgroup of all functions which
map V into S . In particular, if S = {0}, then Φ(V ) is the subgroup of all functions
vanishing at every point of V .
Similarly, suppose that R is a commutative ring and S is an ideal of R. Then
Map(X, R) is a commutative ring under pointwise addition and multiplication.
Further suppose that Y is a subring of Map(X, R). Then for each V ⊂ X , Φ(V ) is
the set of all functions f in F which map V into S : this is an ideal of Y .
Subexample 2.1: K a ﬁeld, X = K n , R = K , S = 0, F = K [x1 , . . . , xn ]. This is
the Galois correspondence between ideals in K [x1 , . . . , xn ] and aﬃne subvarieties
of K n . It will be discussed in more detail in §X.X .
Subexample 2.2: X a compact topological space, R = R, S = 0, Y the set of
all continuous functions from X to R. This will be discussed in more detail in
§X.X . 20 PETE L. CLARK Example 3: Let V be a K vector space and ⟨, ⟩ a bilinear form on K . Let
X = Y = 2V , and deﬁne a Galois connection (Φ, Ψ) by S ⊂ X → S ⊥ = {x ∈
X  ∀s ∈ S ⟨s, x⟩ = 0} and T ⊂ Y → ⊥ T = {x ∈ X  ∀t ∈ T, ⟨x, t⟩ = 0}. Note that
if the bilinear form is symmetric than this is an example of a Galois connection
with Φ = Ψ.
Subexample 3.1: Let ⟨, ⟩ be a symmetric bilinear form on a ﬁnite dimensional
vector space V over a ﬁeld K of characteristic diﬀerent from 2. Then every subspace of V is closed iﬀ {0} is closed iﬀ the form is nondegenerate.
Example 4: Let L be a language, let X be the set of Ltheories, and let Y be
the class of all classes C of Lstructures, each partially ordered by inclusion.7 For
a theory T , we deﬁne Φ(T ) = CT to be the class of all models of T , whereas for
a class C , we deﬁne Ψ(C ) to be the collection of all sentences φ such that for all
X ∈ C , X = φ.
Exercise X.X (Covering Spaces): Let π : X ′ → X be a covering map of pathconnected topological spaces. Let G = Deck(X ′ /X ) the be the group of homeomorphisms σ : X ′ → X ′ such that π ◦ σ = π .
a) Deﬁne a Galois connection between the subgroups of G and the subcovering
spaces πY : Y → X of π . b) What is the induced Galois correspondence?
c) How does the situation simplify when X ′ is the universal cover of X ?
2.3. Examples of Isotone Galois Connections.
COMPLETE ME! ♣
3. Modules
3.1. Basic deﬁnitions.
Suppose (M, +) is an abelian group. For any m ∈ M and any integer n, one
can make sense of n • m. If n is a positive integer, this means m + · · · + m (n times);
if n = 0 it means 0, and if n is negative, then n • m = −(−n) • m. Thus we have
deﬁned a function • : Z × M → M which enjoys the following properties: for all
n, n1 , n2 ∈ Z, m, m1 , m2 ∈ M , we have
(ZMOD1)
(ZMOD2)
(ZMOD3)
(ZMOD4) 1 • m = m.
n • (m1 + m2 ) = n • m1 + n • m2 .
(n1 + n2 ) • m = n1 • m + n2 • m.
(n1 n2 ) • m = n1 • (n2 • m) It should be clear that this is some kind of ringtheoretic analogue of a group
action on a set. In fact, consider the slightly more general construction of a monoid
(M, ·) acting on a set S : that is, for all n1 , n2 ∈ M and s ∈ S , we require 1 • s = s
and (n1 n2 ) • s = n1 • (n2 • s).
7Here we are cheating a bit by taking instead of a partially ordered set, a partially ordered
class. We leave it to the interested reader to devise a remedy. COMMUTATIVE ALGEBRA 21 For a group action G on S , each function g • : S → S is a bijection. For monoidal
actions, this need not hold for all elements: e.g. taking the natural multiplication
action of M = (Z, ·) on S = Z, we ﬁnd that 0• : Z → {0} is neither injective nor
surjective, ±1• : Z → Z is bijective, and for n > 1, n• : Z → Z is injective but not
surjective.
Exercise: Let • : M × S → S be a monoidal action on a set. Show that for
each unit m ∈ M – i.e., an element for which there exists m′ with mm′ = m′ m = 1
– m• : S → S is a bijection.
Then the above “action” of Z on an abelian group M is in particular a monoidal
action of (Z, ·) on the set M . But it is more: M has an additive structure, and
(ZMOD2) asserts that for each n ∈ Z, n• respects this structure – i.e., is a homomorphism of groups; also (ZMOD3) is a compatibility between the additive
structure on Z and the additive structure on M .
These axioms can be restated in a much more compact form. For an abelian group
M , an endomorphism of M is just a group homomorphism from M to itself:
f : M → M . We write End(M ) for the set of all endomorphisms of M . But End(M )
has lots of additional structure: for f, g ∈ End(M ) we deﬁne f + g ∈ End(M ) by
(f + g )(m) := f (m) + g (m),
i.e., pointwise addition. We can also deﬁne f · g ∈ End(M ) by composition:
(f · g )(m) := f (g (m)).
Proposition 9. For any abelian group M , the set End(M ) of group endomorphisms
of M , endowed with pointwise addition and multiplication by composition, has the
structure of a ring.
Exercise: Prove Proposition X.X.
Exercise: Show that End(Z) = Z, and for any n ∈ Z, End(Z/nZ) = Z/nZ. (More
precisely, ﬁnd canonical isomorphisms.)
These simple examples are potentially misleading: we did not say that the multiplication was commutative, and of course there is no reason to expect composition
of functions to be commutative.
Exercise X.X: a) Show that End(Z/2Z ⊕ Z/2Z) = M2 (Z/2Z), the (noncommutative!) ring of 2 × 2 matrices with Z/2Zcoeﬃcients.
b) If M is an abelian group and n ∈ Z+ , show End(M n ) = Mn (End(M )).
Now observe that the statement that the action of Z on M satisfes (ZMOD1)
through (ZMOD4) is equivalent to the following much more succinct statement:
For any abelian group M , the map n ∈ Z → (n•) : M → M is a homomorphism of rings Z → End(M ).
This generalizes very cleanly: if R is any ring (not necessarily commuative) and
M is an abelian group, a homomorphism • : R → End(M ) will satisfy: for all 22 PETE L. CLARK r ∈ R, m, m1 , m2 ∈ M :
(LRMOD1)
(LRMOD2)
(LRMOD3)
(LRMOD4) 1 • m = m.
r • (m1 + m2 ) = r • m1 + r • m2 .
(r1 + r2 ) • m = r1 • m + r2 • m.
(r1 r2 ) • m = r1 • (r2 • m). The terminology here is that such a homomorphism r → (r•) is a left Rmodule
structure on the abelian group M .
What then is a right Rmodule structure on M ? The pithy version is that
it is a ring homomorphism from Rop , the opposite ring of R to End(M ). This deﬁnition makes clear (only?) that if R is commutative, there is no diﬀerence between
left and right Rmodule structures. Since our interest is in the commutative case,
we may therefore not worry too much. But for the record:
Exercise X.X.X: Show that a homomorphism Rop → End(M ) is equivalent to a
mapping • : M × R → M satisfying
m • 1 = m,
(m1 + m2 ) • r = m1 • r + m2 • r,
m • (r1 + r2 ) = m • r1 + m • r2 ,
m • (r1 r2 ) = (m • r1 ) • r2 . As usual for multiplicative notation, we will generally suppress the bullet, writing rm for left Rmodules and mr for right Rmodules.
The calculus of left and right actions is at the same time confusing and somewhat miraculous: it is a somewhat disturbing example of a purely lexicographical
convention that has – or looks like it has – actual mathematical content. Especially, suppose we have an abelian group M and two rings R and S , such that M
simultaneously has the structure of a left Rmodule and a right S module. Thus
we wish to entertain expressions such as rms for m ∈ M , r ∈ R, s ∈ S . But as
stands this expression is ambiguous: it could mean either
(r • m) • s
or
r • (m • s).
We say that M is an RS bimodule if both of these expressions agree. Here is what
is strange about this: lexicographically, it is an associativity condition. But “really” it is a commutativity condition: it expresses the fact that for all r ∈ R, s ∈ S ,
(r•) ◦ (•s) = (•s) ◦ (r•): every endomorphism coming from an element of R commutes with every endomorphism coming from an element of S . Thus for instance:
Exercise: Show that any ring R is naturally a left Rmodule and a right Rmodule.
We will not deal with bimodules further in these notes. In fact, when we say COMMUTATIVE ALGEBRA 23 Rmodule at all, it will be understood to mean a left Rmodule, and again, since
we shall only be talking about commutative rings soon enough, the distinction between left and right need not be made at all.
Deﬁnition: For M a left Rmodule, we deﬁne its annihilator
ann(M ) = {r ∈ R  ∀m ∈ M, rm = 0}.
Equivalently, ann(M ) is the set of all r such that r· = 0 ∈ End(M ), so that it is
precisely the kernel of the associated ring homomorphism R → End(M ). It follows
that ann(M ) is an ideal of R (note: twosided, in the noncommutative case).
Deﬁnition: A left Rmodule M is faithful if ann(M ) = 0. Explicitly, this means
that for all 0 ̸= r ∈ R, there exists m ∈ M such that rm ̸= 0.
Exercise X.X: Let M be an Rmodule. Show that M has the natural structure
of a faithful R/ ann(M )module.
Deﬁnition: Let M be a left Rmodule. A submodule of M is a subgroup N
of (M, +) such that RN ⊂ N . The following result is extremely easy and allimportant:
Theorem 10. Let R be a ring. The left Rsubmodules of R are precisely the left
ideals of R.
Exercise X.X: Prove Theorem 10.
Deﬁnition: Let M and N be left Rmodules. A homomorphism of Rmodules
is a homomorphism of abelian groups f : M → N such that for all r ∈ R, m ∈
M, n ∈ N , f (rm) = rf (m).
Exercise X.X: a) Deﬁne an isomorphism of Rmodules in the correct way, i.e.,
not as a bijective homomorphism of Rmodules.
b) Show that a homomorphism of Rmodules is an isomorphism iﬀ it is bijective.
If N is a submodule of a left Rmodule M , then the quotient group M/N has
a natural Rmodule structure. More precisely, there is a unique left Rmodule
structure on M/N such that the quotient map M → M/N is a homomorphism of
Rmodules. (Exercise!)
Exercise: Let I be a twosided ideal of the ring R, so that the quotient ring R/I
has the structure of a left Rmodule. Show that
ann(R/I ) = I.
In particular, every twosided ideal of R occurs as the annihilator of a left Rmodule.
Exercise X.X: a) Let⊕ be a ring and {Mi }i∈I a family of Rmodules. Consider the
R
abelian group M = i∈I Mi . Show that putting r(mi ) = (rmi ) makes R into an
Rmodule. Show that the usual inclusion map ιi : Mi → M is a homomorphism of
Rmodules.
b) Show that for any Rmodule N and Rmodule maps fi : Mi → N , there exists 24 PETE L. CLARK a unique Rmodule map f : M → N such that fi = f ◦ ιi for all i ∈ I . Thus M
satisﬁes the universal mapping property of the direct sum.
As a matter of notation, for n ∈ Z+ , Rn := ⊕n
i=1 R, R0 = 0. Exercise X.X: Work out the analogue of Exercise X.X for direct products.
Exercise X.X: a) Suppose that M is an Rmodule and S is a subset of M . Show
that the intersection of all Rsubmodules of M containing S is an Rsubmodule,
and is contained in every Rsubmodule that contains S . We call it the Rsubmodule
generated by S .
b)
∑ If S = {si }i∈I , show that the Rmodule generated by S is the set of all sums
i∈J ri si , where J is a ﬁnite subset of S .
Exercise X.X: Suppose that k is a ﬁeld. Show that the terms “k module” and
“vector space over k ” are synonymous.
One can therefore view the theory of Rmodules as a generalization of vector spaces
to arbitrary rings. But really this is something like a zeroth order approximation
of the truth: for a general ring R, the theory of Rmodules is incomparably richer
than the theory of vector spaces over a ﬁeld. There are two explanations for this.
First, even when working with very simple Rmodules such as Rn , the usual linear
algebra notions of linear independence, span and basis remain meaningful, but behave in unfamiliar ways:
Call a subset S of an Rmodule M linearly independent if for every ﬁnite
subset m1 , . . . , mn of S and any r1 , . . . , rn ∈ R, r1 m1 + . . . + rn mn = 0 implies
r1 = . . . = rn = 0. Say that S spans R if the Rsubmodule generated by S is R,
and ﬁnally a basis for an Rmodule is a subset which is both linearly independent
⊕
and spanning. For example, for any set I , the Rmodule i R has a basis ei .
In linear algebra – i.e., when R is a ﬁeld – every Rmodule has a basis.8 However the situation is quite diﬀerent over a general ring:
Theorem 11. a) Let M be an Rmodule. Suppose that S ⊂ R is a basis. Then M
⊕
is isomorphic as an Rmodule to s∈S R.
⊕
b) Let S ⊕ any set, and consider the Rmodule RS := s∈S R. For each s ∈ S ,
be
let es ∈
s∈S R be the element whose scoordinate is 1 and all of whose other
coordinates are 0. Then set {es }s∈S is a basis for RS .
Exercise X.X: Prove Theorem X.X.
A module which has a basis – so, by the theorem, admits an isomorphism to
for some index set S – is called free. ⊕
s∈S R Exercise X.X: Show that a nonzero free Rmodule is faithful.
8This uses, and is in fact equivalent to, the Axiom of Choice, but the special case that any
vector space with a ﬁnite spanning set has a basis does not. COMMUTATIVE ALGEBRA 25 Let us examine the case of modules over R = Z, i.e., of abelian groups. In this case,
the term free abelian group is synonmyous with free Zmodule. Probably the
reader knows that not all abelian groups are free, but in case not: for any integer
n > 1, the Zmodule Z/nZ is not free, since it has nonzero annihilator nZ. Thus
Z/nZ does not have a basis as a Zmodule, and indeed has no nonempty linearly
independent subsets!
In fact:
Proposition 12. For a commutative ring R, TFAE:
(i) Every Rmodule is free.
(ii) R is a ﬁeld.
Proof. As discussed above, (ii) =⇒ (i) is a fundamental theorem of linear algebra,
so we need only concern ourselves with the converse. But if R is not a ﬁeld, then
(Exercise X.X) there exists a nonzero proper ideal I , and then R/I is a nontrivial
Rmodule with 0 ̸= I = ann(R/I ), so by Exercise X.X R/I is not free.
Noncommutative Remark: If R is a noncommutative ring such that every left Rmodule is free, then the above argument shows R has no nonzero proper twosided
ideals, so is what is called a simple ring. But a noncommutative simple ring may
still admit a nonfree module. For instance, let k be a ﬁeld and take R = M2 (k ),
the 2 × 2 matrix ring over k . Then k ⊕ k is a left Rmodule which is not free. However, suppose R is a ring with no proper nontrivial onesided ideals. Then R is a
division ring – i.e., every nonzero element of R is a unit – and every Rmodule is free.
In linear algebra – i.e., when R is a ﬁeld – every linearly independent subset of
an Rmodule can be extended to a basis. Over a general ring this does not hold
even for free Rmodules. For instance, take R = M = Z. A moment’s thought
reveals that the only two bases are {1} and {−1}, whereas the linearly independent
sets are precisely the singleton sets {n} as n ranges over the nonzero integers.
Final Remark: Note well the form of Proposition 12: we assume that R is a
commutative ring for which all Rmodules (or, more generally, some nice class
of Rmodules) satisfy some nice property, and we deduce a result on the structure
of R. Such problems – roughly, “inverse problems” – have a broad aesthetic appeal
throughout mathematics, and provide one of the major motivations for studying
modules above and beyond their linearalgebraic origins. We will see other such
characterizations later on.
3.2. Finitely presented modules.
One of the major diﬀerences between abelian groups and nonabelian groups is
that a subgroup N of a ﬁnitely generated abelian group M remains ﬁnitely generated, and indeed, the minimal number of generators of the subgroup N cannot
exceed the minimal number of generators of M , whereas this is not true for nonabelian groups: e.g. the free group of rank 2 has as subgroups free groups of every
rank 0 ≤ r ≤ ℵ0 . (For instance, the commutator subgroup is not ﬁnitely generated.)
Since an abelian group is a Zmodule and every Rmodule has an underlying abelian
group structure, one might well expect the situation for Rmodules to be similar to 26 PETE L. CLARK that of abelian groups. We will see later that this is true in many but not all cases:
an Rmodule is called Noetherian if all of its submodules are ﬁnitely generated.
Certainly a Noetherian module is itself ﬁnitely generated. The basic fact here –
which we will not prove until §X.X , despite its relative simplicity – is a partial
converse: if the ring R is Noetherian, any ﬁnitely generated Rmodule is Noetherian. Note that we can already see that the Noetherianity of R is necessary: if R is
not Noetherian, then by deﬁnition there exists an ideal I of R which is not ﬁnitely
generated, and this is nothing else than a nonﬁnitely generated Rsubmodule of R
(which is itself generated by the single element 1.) Thus the aforementioned fact
about subgroups of ﬁnitely generated abelian groups being ﬁnitely generated holds
because Z is a Noetherian ring.
When R is not Noetherian, it becomes necessary to impose stronger conditions
than ﬁnite generation on modules. One such condition indeed comes from group
theory: recall that a group G is ﬁnitely presented if it is isomorphic to the quotient of a ﬁnitely generated free group F by the least normal subgroup N generated
by a ﬁnite subset x1 , . . . , xm of F .
Proposition 13. For a ﬁnitely generated Rmodule M , TFAE:
(i) There exist nonnegative integers m, n and an exact sequence
R m → R n → M → 0.
(ii) M is the quotient of an f.g. free Rmodule Rn by some f.g. submodule N .
A module M satisfying these equivalent conditions is said to be ﬁnitely presented.
Proof. That (i) implies (ii) is immediate. Conversely, let M = Rn /N where N is
ﬁnitely generated. Then there exists a surjection Rm → N and thus the sequence
Rm → Rn → M → 0
is exact.
Proposition 14. Let
ψ ϕ 0→K→N →M →0
be a short exact sequence of Rmodules, with M ﬁnitely presented and N ﬁnitely
generated. Then K is ﬁnitely generated.
Proof. (Matsumura) By deﬁnition of ﬁnitely presented, we can place M in an exact
sequence
(1) f Rm → Rn M → 0 for some m, n ∈ N. For 1 ≤ i ≤ n, let ei be the ith standard basis element of M ,
let mi = f (ei ) be the image in M , and choose ni ∈ N any element in ϕ−1 (mi ).
Then there is a unique Rmodule homomorphism α : Rn → N given by α(ei ) = ni ,
which restricts to an Rmodule homomorphism β : B m → K . Altogether we get a
commutative diagram
f Rm −→ Rn −→ M −→ 0
ψ ϕ 0 −→ K −→ N −→ M. COMMUTATIVE ALGEBRA 27 The rest of the proof is essentially a diagram chase. Suppose N = ⟨ξ1 , . . . , ξk ⟩R ,
and choose v1 , . . . , vk ∈ Rn such that ϕ(ξi ) = f (vi ). Put
′
ξi = ξi − α(vi ).
′
Then φ(ξi ) = 0, so there exist unique ηi ∈ K such that
′
ξi = ψ (ηi ). We claim that K is generated as an Rmodule by β (Rm ) and η1 , . . . , ηk and thus
is ﬁnitely generated. Indeed, for η ∈ K , there are r1 , . . . , rk ∈ R such that
∑
ψ (η ) =
ri ξi .
i Then
ψ (η − ∑ ri ηi ) = i Since we may write ∑ ∑
′
ri (ξi − ξi ) = α(
ri vi ). i i ∑
∑
0 = ϕ(α(
ri vi )) = f (
ri vi ),
i ∑ i = g (u) with u ∈ Rm . Then
∑
∑
ψ (β (u)) = α(g (u)) = α(
ri vi ) = ψ (η −
ri ηi ).
i ri vi i Since ψ is injective, we conclude
η = β ( u) + i ∑ ri ηi . i Exercise X.X:
Let 0 → M ′ → M → M ′′ → 0 be a short exact sequence of Rmodules.
a) Show that if M ′ and M ′′ are both ﬁnitely presented, so is M .
b) Show that if M is ﬁnitely presented and M ′ is ﬁnitely generated, then M ′′ is
ﬁnitely presented.
A stronger condition yet is the following: an Rmodule M is coherent if it is
ﬁnitely generated and every ﬁnitely generated submodule is ﬁnitely presented. Evidently coherent implies ﬁnitely presented implies ﬁnitely generated, and all three
coincide over a Noetherian ring. The signiﬁcance of coherence lies in the following:
Theorem 15. Let R be a ring (not necessarily commutative, but with unity).
a) The category of all left Rmodules is an abelian category.
b) The category of all coherent left Rmodules is an abelian category.
c) In particular, if R is left Noetherian, the category of all ﬁnitely generated left
Rmodules is an abelian category.
d) There exists a commutative ring R for which the category of all ﬁnitely generated
(left) Rmodules is not abelian.
The proof of this result – and even an explanation of the term “abelian category”
is beyond the scope of these notes, so this is deﬁnitely an ancillary remark. Nevertheless we hope that it will be of some use to students of algebraic geometry:
for instance, it explains the fact that Hartshorne only deﬁnes coherent sheaves of 28 PETE L. CLARK modules on a scheme X in the case that the scheme is Noetherian and suggests the
correct (and more subtle) deﬁnition in the nonNoetherian case.
3.3. Torsion and torsionfree modules.
Let R be a domain, and let M be an Rmodule. An element x ∈ M is said to
be torsion if there exists 0 ̸= a ∈ R such that ax = 0. Equivalently, the annihilator ann(x) = {a ∈ R  ax = 0} is a nonzero ideal of R. We deﬁne M [tors] to
be the set of all torsion elements of M . It is immediate to see that M [tors] is a
submodule of M . We say that M is a torsion Rmodule if M = M [tors] and that
M is torsionfree if M [tors] = 0.
Exercise: Let 0 → M1 → M → M2 → 0 be an exact sequence of modules.
a) Show that if M is torsion, so are M1 and M2 .
b) If M1 and M2 are torsion modules, must M be torsion?
c) Show that if M is torsionfree, show that so is M1 , but M2 need not be.
d) If M1 and M2 are torsionfree, must M be torsionfree?
Proposition 16. Let R be an integral domain and M an Rmodule.
a) The quotient M/M [tors] is torsionfree.
b) If M is ﬁnitely generated and torsionfree, then it is isomorphic to a submodule
of a ﬁnitely generated free module.
Proof. a) Put N = M/M [tors], and let x ∈ N be such that there exists 0 ̸= a ∈ R
with ax = 0. Let x be any lift of x to M ; then there exists t ∈ M [tors] such
that ax = t. By deﬁnition of torsion, there exists a′ ∈ R such that a′ t = 0, so
a′ ax = a′ t = 0. Since R is a domain, a′ a is nonzero, so x ∈ M [tors] and x = 0.
c) Suppose M = ⟨x1 , . . . , xr ⟩. We may of course assume that M ̸= 0 and hence
that r ≥ 1 and all the xi are nonzero. Further, after reordering the xi ’s if necessary,
there exists a unique s, 1 ≤ s ≤ r, such that {x1 , . . . , xs } is linearly independent
over R but for all i with s < i ≤ r, {x1 , . . . xs , xi } is linearly dependent over R.
Then F = ⟨x1 , . . . , xs ⟩ ∼ Rs , so we are done if s = r. If s < ∏ then for each
r,
=
i > s there exists 0 ̸= ai ∈ R such that ai xi ∈ F . Put a = s<i≤r ai : then
aM ⊂ F . Let [a] : M → M denote multiplication by a. Since M is torsionfree, [a]
is injective hence gives an Rmodule isomorphism from M to a submodule of the
ﬁnitely generated free module F .
Exercise: Show that the torsionfree Zmodule (Q, +) is not isomorphic to a submodule of any ﬁnitely generated free Zmodule. Thus – even for very nice rings! –
the hypothesis of ﬁnite generation is necessary in Proposition 16. 3.4. Tensor and Hom.
3.4.1. Tensor products.
We assume that the reader has some prior familiarity with tensor products, say
of vector spaces and/or of abelian groups. The ﬁrst is an instance of tensor products of k modules, for some ﬁeld k , and the second is an instance of tensor products
of Zmodules. We want to give a general deﬁnition of M ⊗R N , where M and N COMMUTATIVE ALGEBRA 29 are two Rmodules.
There are two ways to view the tensor product construction: as a solution to a
universal mapping problem, and as a generators and relations construction. They
are quite complementary, so it is a matter of taste as to which one takes as “the”
deﬁnition. So we will follow our taste by introducing the mapping problem ﬁrst:
Suppose M , N , P are Rmodules. By an Rbilinear map f : M × N → P we
mean a function which is separately Rlinear in each variable: for all m ∈ M , the
mapping n → f (m, n) is Rlinear, and for each n ∈ N , the mapping m → f (m, n) is
Rlinear. Now consider all pairs (T, ι), where T is an Rmodule and ι : M × N → T
is an Rbilinear map. A morphism from (T, ι) to (T ′ , ι′ ) will be an Rmodule homomorphism h : T → T ′ such that ι′ = h ◦ ι. By deﬁnition, a tensor product M ⊗R N
is an initial object in this category: i.e., it comes equipped with an Rbilinear map
M × N raM ⊗R N such that any Rbilinear map f : M × N → P factors through
it. As usual, the initial object of a category is unique up to unique isomorphism
provided it exists.
As for the existence, we fall back on the generators and relations construction.
Namely, we begin with the free Rmodule F whose basis is M × N , and we write
the basis elements (purely formally) as m ⊗ n. We then take the quotient by the
submodule generated by the following relations R:
(x + x′ ) ⊗ y − x ⊗ y − x′ ⊗ y,
x ⊗ (y + y ′ ) − x ⊗ y − x ⊗ y ′ ,
(ax) ⊗ y − a(x ⊗ y ),
x ⊗ (ay ) − a(x ⊗ y ).
It is then easy to see that the quotient map M × N → F/N satisﬁes all the properties of a tensor product (details left to the reader).
Note that the general element of M ⊗R N is not a single element of the form x ⊗ y
but rather a ﬁnite sum of such elements. (Indeed, from the free Rmodule, every
element can be represented by a ﬁnite Rlinear combination of elements of the form
x ⊗ y , but the last two deﬁning relations in the tensor product allow us to change
ri (x ⊗ y ) to either (ri x) ⊗ y or x ⊗ (ri y ).) Of course, this representation of an element
of the tensor product need not be (and will never be, except in trivial cases) unique.
One can also take the tensor product of Ralgebras: if R is a (commutative!) ring
and A and B are commutative Ralgebras, then on the tensor product A ⊗R B we
have a naturally deﬁned product, induced by (a1 ⊗ b1 ) · (a2 ⊗ b2 ) := (a1 a2 ⊗ b1 b2 ).
We have to check that this is welldeﬁned, a task which we leave to the reader (or
see [AM, pp. 3031])). The tensor product of algebras is a powerful tool – e.g.
in the structure theory of ﬁnitedimensional algebras over a ﬁeld, or in the theory
of linear disjointness of ﬁeld extensions – and is given misleadingly short shrift in
most elementary treatments.
Base change: Suppose that M is an Rmodule and f : R → S is a ring homomorphism. Then S is in particular an Rmodule, so that we can form the tensor
product S ⊗R M . This is still an Rmodule, but it is also an S module in an evident 30 PETE L. CLARK ∑
∑
way: s • ( i si ⊗ mi ) := i ssi ⊗ mi . This is process is variously called scalar
extension, base extension or base change. Note that this process is functorial, in the following sense: if f : M → M ′ is an Ralgebra homomorphism, then
there exists an induced S algebra homomorphism S ⊗R M → S ⊗R M ′ , given by
s ⊗ m → s ⊗ f (m).
Exercise X.X: If M is a ﬁnitely generated Rmodule and f : R → S is a ring
homomorphism, then S ⊗R M is a ﬁnitely generated S module.
Exercise X.X: Let A and B be rings, M an Amodule, P a B module, and N
an (A, B )bimodule. Then M ⊗A N is naturally a B module, N ⊗B P is naturally
an Amodule, and
(M ⊗A N ) ⊗B P ∼ M ⊗A (N ⊗B P ).
=
Exercise X.X: Let R be a commutative ring, I and ideal of R and M an Rmodule.
a) Show that there is a welldeﬁned Rbilinear map R/I × M → M/IM given by
(r + I, m) → rm + I . Thus there is an induced homomorphism of Rmodules
φ : R/I ⊗R M → M/IM.
b) Show that φ is an isomorphism of Rmodules.
Proposition 17. Let R be a commutative ring, M an Rmodule and {Ni }i∈I a
directed system of Rmodules. Then the Rmodules lim(M ⊗ Ni ) and M ⊗ (lim Ni )
−
→
−
→
are canonically isomorphic.
Exercise X.X: Prove Proposition 17.
3.4.2. Hom modules.
3.4.3. Compatibilities between ⊗ and Hom.
3.5. Projective modules.
3.5.1. Basic equivalences.
Proposition 18. For an Rmodule P , TFAE:
(i) There exists an Rmodule Q such that P ⊕ Q is a free Rmodule.
(ii) If π : M → N is a surjective Rmodule homomorphism and φ : P → N is a
homomorphism, then there exists at least one Rmodule homomorphism Φ : P → M
such that φ = π ◦ Φ.
(iii) If π : M → N is a surjection, then the natural map Hom(P, M ) → Hom(P, N )
given by Φ → π ◦ Φ is surjective.
(iv) The functor Hom(P, ) is exact.
(v) Any short exact sequence of Rmodules
q 0→N →M →P →0
splits: there exists an Rmodule map σ : P → M such that q ◦ σ = 1P and thus an
internal direct sum decomposition M = N ⊕ σ (P ).
A module satisfying these equivalent conditions is called projective. COMMUTATIVE ALGEBRA 31 Proof. (i) =⇒ (ii): Let F ∼ P ⊕ Q be a free module. Let {fi } be a free basis for
=
F and let {pi } be the corresponding generating set for P , where pi is the image of
fi under the natural projection P ⊕ Q → P . Put ni = φ(pi ). By surjectivity of π ,
let mi ∈ π −1 (ni ). By the freeness of F , there is a unique Rmodule homomorphism
h : F → M carrying each fi to mi . Pull h back to P via the natural inclusion
P → F . Then h : P → M is such that π ◦ f = φ.
(ii) =⇒ (i): As for any Rmodule, there exists a free Rmodule F and a surjection
π : F → P . Applying (ii) with N = p and φ : P → N the identity map, we get a
homomorphism Φ : P → F such that π ◦ φ = 1P . It follows that F = Φ(P ) ⊕ ker(π )
is an internal direct sum decomposition.
(ii) ⇐⇒ (iii): (iii) is nothing more than a restatement of (ii), as we leave it to the
reader to check.
(iii) ⇐⇒ (iv): To spell out (iv), it says: if
0 → M ′ → M → M ′′ → 0
is a short exact sequence of Rmodules, then the corresponding sequence
0 → Hom(P, M ′ ) → Hom(P, M ) → Hom(P, M ′′ ) → 0
is exact. It is easy to see that for any Rmodule P , the sequence
0 → Hom(P, M ′ ) → Hom(P, M ) → Hom(P, M ′′ )
is exact – i.e., Hom(P, ) is left exact – so (iv) amounts to: for any surjection
M → M ′′ , the corresponding map Hom(P, M ) → Hom(P, M ′′ ) is surjective, and
this is condition (iii).
(ii) =⇒ (v): Given
q
0 → N → M → P → 0,
we apply (ii) to the identity map 1P : P → P and the surjection q : M → P ,
getting a map σ : P → M such that q ◦ σ = 1P , so σ is a section as required.
(v) =⇒ (i): Choosing a set of generators for P gives rise to a surjective homomorphism q : F → P from a free Rmodule F to P and thus a short exact
sequence
q
0 → Ker q → F → P → 0.
By hypothesis, there exists a section σ : P → F and thus an internal direct sum
decomposition F ∼ Ker(q ) ⊕ σ (P ) ∼ Ker(q ) ⊕ P .
=
=
Exercise X.X: Give a direct proof that (v) =⇒ (ii) in Proposition 18. (Suggestion:
Given the surjection q : M → N and the map π : P → N , form the short exact
sequence 0 → K → M → N → 0 and show that it is mapped to by a short exact
sequence 0 → K → M ×N P → P → 0, where
M ×N P = {(x, y ) ∈ M × P  q (x) = π (y )}
is the ﬁber product of M and P over N .)
Exercise X.X: Use Proposition 18 to show, several times over, that a free Rmodule
is projective.
Exercise X.X: Let {Mi }i∈I be an index family of Rmodules. Show that the direct
⊕
sum M = i∈I Mi is projective iﬀ each Mi is projective. 32 PETE L. CLARK Exercise X.X:
a) Show that the tensor product of two free Rmodules is free.
b) Show that the tensor product of two projective Rmodules is projective.
Exercise X.X: Show that a ﬁnitely generated projective module is ﬁnitely presented.
(Hint: the problem is that over a notnecessarilyNoetherian ring, a submodule of
a ﬁnitely generated module need not be ﬁnitely generated. However, a direct summand of a ﬁnitely generated module is always ﬁnitely generated: why?)
3.5.2. Linear algebraic characterization of projective modules.
Let R be a commutative ring, n ∈ Z+ , and let P be an element of the (noncommutative!) ring Mn (R) of n × n matrices with entries in R such that P 2 = P .
There are several times for such a matrix. The pure algebraist would call such a
matrix idempotent, for that is the name of an element in any ring which is equal
to its square. A geometrically minded algebraist however may call such a matrix
a projection, the idea being that the corresponding Rmodule endomorphism of
Rn “projects” Rn onto the submodule P (Rn ).
Proposition 19. An Rmodule M is ﬁnitely generated and projective iﬀ it is, up
to isomorphism, the image of a projection: i.e., iﬀ there exists n ∈ Z+ and a matrix
P ∈ Mn (R) with P = P 2 such that M ∼ P (Rn ).
=
Proof. Suppose ﬁrst that M is a ﬁnitely generated projective Rmodule. Since M
is ﬁnitely generated,, there exists n ∈ Z+ and a surjective Rmodule homorphism
π : Rn → M . Since M is projective, this homomorphism has a section σ : M → Rn ,
and we may thus write Rn = σ (M ) ⊕ M ′ . Put P = σ ◦ π ∈ EndR (Rn ). Then
P (Rn ) = σ (π (Rn )) = σ (M ) ∼ M and
=
P 2 = σ ◦ (π ◦ σ ) ◦ π = σ ◦ 1M ◦ π = σ ◦ π = P.
Conversely, suppose that there exists P ∈ EndR (Rn ) with P 2 = P and let M ∼
=
P (Rn ). Then – since P (1 − P ) = 0 – Rn = P (Rn ) ⊕ (1 − P )(Rn ), exhibiting P (Rn )
as a direct summand of a free module.9
3.5.3. The Dual Basis Lemma.
Proposition 20. (Dual Basis Lemma) For an Rmodule M , TFAE:
(i) There exists an index set I , elements {ai }i∈I of M and homomorphisms {fi :
M → R}i∈I such that for each a ∈ M , {i ∈ I  fi (a) ̸= 0} is ﬁnite, and
∑
a=
fi (a)ai .
i ∈I (ii) M is projective.
Proof. (i) =⇒ (ii): Let F be the free Rmodule with basis elements {ei }i∈I ,
and deﬁne f : F → M by f (ei ) = ai . Then the map ι : M → F given by
∑
ι(a) = i∈I fi (a)ei is a section of f , so M is a direct summand of F .
⊕
(ii) =⇒ (i): Let f : F = i∈I R → M be an epimorphism from a free Rmodule
9Note that this part of the proof redeems the pure algebraist: this the decomposition aﬀorded
by the pair of orthogonal idempotents P, 1 − P . COMMUTATIVE ALGEBRA 33 onto M . Since M is projective, there exists a section ι : M → F . If {ei }i∈I is the
standard basis of F , then for all a ∈ M , the expression
∑
ι(a) =
fi (a)ei
i ∈I deﬁnes the necessary family of functions fi : M → R.
Exercise: Let P be a projective Rmodule. Show that one can can ﬁnd a ﬁnite index
set I satisfying condition (i) of the Dual Basis Lemma iﬀ P is ﬁnitely generated.
3.5.4. Projective versus free.
Having established some basic facts about projective modules, we should now seek
examples in nature: which modules are projective? Note that by Exercise X.X any
free module is projective. But this surely counts as a not very interesting example!
Indeed the following turns out to be one of the deepest questions of the subject.
Question 1. When is a projective module free?
We want to give examples to show that the answer to Question 1 is not “always”.
But even by giving examples one wades into somewhat deep waters. The following
is the one truly “easy” example of a nonfree projective module I know.
Example: Suppose R1 and R2 are nontrivial rings. Then the product R = R1 × R2
admits nonfree projective modules. Indeed, let P be the ideal R1 × {0} and Q the
ideal {0} × R2 . Since R = P ⊕ Q, P and Q are projective. On the other hand P
cannot be free because taking e := (0, 1) ∈ R, we have eP = 0, whereas eF ̸= 0 for
any nonzero free Rmodule F (and of course, Q is not free either for similar reasons).
One way to construe Question 1 is to ask for the class of rings over which every
projective module is free, or over which every ﬁnitely generated projective module
is free. I actually do not myself know a complete answer to this question, but there
are many interesting and important special cases.
Recall the following result from undergraduate algebra.
Theorem 21. A ﬁnitely generated module over a PID is free iﬀ it is torsionfree.
Of course submodules of torsionfree modules are torsionfree, so projective implies
torsionfree. We deduce:
Corollary 22. A ﬁnitely generated projective module over a PID is free.
Theorem 21 does not extend to all torsionfree modules: for instance, the Zmodule
Q is torsionfree but not free. However Corollary 22 does extend to all modules over
a PID. The proof requires transﬁnite methods and is given in §3.10.
Recall that a ring R is local if it has a unique maximal ideal. It is convenient
to reserve the notation m for the unique maximal ideal of a local ring and speak of
“the local ring (R, m)”. We want to show that every ﬁnitely generated projective
module over a local ring is free. First a few preliminaries. 34 PETE L. CLARK Let f : R → S be a homomorphism of rings. Then necessarily f induces a homomorphism f × : R× → S × on unit groups: if xy = 1, then f (x)f (y ) = f (1) = 1,
so units get mapped to units. But what about the converse: if x ∈ R is such that
f (x) is a unit in S , must x be a unit in R?
It’s a nice idea, but it’s easy to see that this need not be the case. For instance,
let a > 1 be any positive integer. Then a is not a unit of Z, but for each prime
p > a, the image of a in the quotient ring Z/pZ is a unit. Too bad! Let us not give
up so soon: a conjecture may fail, but a deﬁnition cannot: say a homomorphism
f : R → S of rings is unitfaithful if for all x ∈ R, f (x) ∈ S × =⇒ x ∈ R× .
Lemma 23. If (R, m) is a local ring, the quotient map q : R → R/m is unitfaithful.
Proof. An element of any ring is a unit iﬀ it is contained in no maximal ideal, so
in a local ring we have R× = R \ m. Moreover, since m is maximal, R/m is a ﬁeld.
Thus, for x ∈ R,
q (x) ∈ (R/m)× ⇐⇒ x ∈ m ⇐⇒ x ∈ R× .
/ Theorem 24. A ﬁnitely generated projective module over a local ring is free.
Proof. Let P be a ﬁnitely generated projective module over the local ring (R, m).
We may ﬁnd Q and n ∈ Z+ such that P ⊕ Q = Rn . Now tensor with R/m:
we get a direct sum decomposition P/mP ⊕ Q/mQ = (R/m)n . Since R/m is a
ﬁeld, all R/mmodules are free. Choose bases {pi } for P/mP and {qj } for Q/mQ,
and for all i, j , lift each pi to an element pi of P and each qj to an element qj
of Q. Consider the n × n matrix A with coeﬃcients in R whose columns are
p1 , . . . , pa , q1 , . . . , qb . The reduction modulo m of A is a matrix over the ﬁeld R/m
whose columns form a basis for (R/m)n , so its determinant is a unit in (R/m)× .
Since det(M (mod m)) = det(M ) (mod m), Lemma 23 implies that det(M ) ∈ R× ,
i.e., M is invertible. But this means that its columns are linearly independent, so
p1 , . . . , pa , a priori only a generating set for the Rmodule P , is in fact a basis.
Once again, in Section 3.9 this result wll be improved upon: it is a celebrated theorem of Kaplansky that any projective module over a local ring is free.
Finally, we give without proof a celebrated result of H. Bass with a clear moral:
ﬁnitely generated projective modules are usually much more interesting than inﬁnitely generated projective modules. Here let us be sure that by an inﬁnitely
generated Rmodule, we mean an Rmodule which is not ﬁnitely generated.10
Theorem 25. (Bass [Bas63]) Let R be connected (i.e., without nontrivial idempotents) Noetherian ring. Then any inﬁnitely generated projective Rmodule is free.
Much more interesting is an example of a ﬁnitely generated projective, nonfree
module over an integral domain. Probably the ﬁrst such examples come from nonprincipal ideals in rings of integers of number ﬁelds with class number greater than
1. To give such an example with proof of its projectivity this early in the day, we
10A priori it would be reasonable to take “inﬁnitely generated Rmodule” to mean a module
which possesses an inﬁnite generating set, but a moment’s thought shows that an Rmodule has
this property iﬀ it is inﬁnite, so it is more useful to deﬁne “inﬁnitely generated” as we have. COMMUTATIVE ALGEBRA 35 require a little preparation.11
Two ideals I and J in a ring R are comaximal if I + J = R. More generally,
a family {Ii } of ideals in a ring is pairwise comaximal if for all i ̸= j , I + J = R.
Lemma 26. Let I , J , K1 , . . . , Kn be ideals in the ring R. Put K = K1 · · · Kn .
a) We have (I + J )(I ∩ J ) ⊂ IJ .
b) If I and J are comaximal, IJ = I ∩ J .
Proof. a) (I + J )(I ∩ J ) = I (I ∩ J ) + J (I ∩ J ) ⊂ IJ + IJ = IJ .
b) If I + J = R, the identity of part a) becomes I ∩ J ⊂ IJ . Since the converse
inclusion is valid for all I and J , the conclusion follows.
Proposition 27. Let I and J be comaximal ideals in a domain R, and consider
the Rmodule map q : I ⊕ J → R given by (x, y ) → x + y . Then:
a) The map q is surjective.
b) Ker(q ) = {(x, −x)  x ∈ I ∩ J }, hence is isomorphic as an Rmodule to I ∩ J .
c) We have an isomorphism of Rmodules
I ⊕ J ∼ IJ ⊕ R.
=
d) Thus if IJ is a principal ideal, I and J are projective modules.
Proof. It is clear that for any ideals I and J , the image of the map q is the ideal
I + J , and we are assuming I + J = R, whence part a).
Part b) is essentially immediate: details are left to the reader.
Combining parts a) and b) we get a short exact sequence
0 → I ∩ J → I ⊕ J → R → 0.
But R is free, hence projective, and thus the sequence splits, giving part c). Finally,
a nonzero principal ideal (x) in a domain R is isomorphic as an Rmodule to R
itself: indeed, multiplication by x gives the isomorphism R → (x). So if IJ is
principal, I ⊕ J ∼ R2 and I and J are both direct summands of a free module.
=
In particular, if we can ﬁnd in a domain R two comaximal nonprincipal ideals I
and J with IJ principal, then I and J are ﬁnitely generated projective nonfree
Rmodules. The following exercise asks you to work through an explicit example.
√
Exercise X.X: Let R = Z[ −5], and put
√
√
p1 = ⟨3, 1 + −5⟩, p2 = ⟨3, 1 − −5⟩.
a) Show that R/p1 ∼ R/p2 ∼ Z/3Z, so p1 and p2 are maximal ideals of R.
=
=
b) Show that p1 + p2 = R (or equivalently, that p1 ̸= p2 ).
c) Show that p1 p2 = (3).
d) Show that neither p1 nor p2 is principal.
√
√
(Suggestion: show that if p1 = (x + −5y ) then p2 = (x − −5y ) and thus there
are integers x, y such that x2 + 5y 2 = ±3.)
e) Conclude that p1 and p2 are (in fact isomorphic) nonfree ﬁnitely generated projective modules over the domain R.
11Here we wish to acknowledge our indebtedness to K. Conrad:
we took our
inspiration for Proposition 27 and the following Exercise from Example 3.1 of
http://www.math.uconn.edu/∼kconrad/blurbs/linmultialg/splittingmodules.pdf. 36 PETE L. CLARK f) Show that p2 is principal, and thus that the class of p in K0 (R) is 2torsion.
This construction looks very speciﬁc, and the numbertheoretically inclined reader
is warmly invited to play around with other quadratic rings and more general rings
of integers of number ﬁelds to try to ﬁgure out what is really going on. From our
perspective, we will (much later on) gain a deeper understanding of this in terms
of the concepts of invertible ideals, the Picard group and Dedekind domains.
Example: Let X be a compact space, and let C (X ) be the ring of continuous
realvalued functions on X . The basic structure of these rings is studied in §X.X .
Let E → X be a real topological vector bundle over X . Then the group Γ(E )
of global sections is naturally a module over C (X ). In fact it is a ﬁnitely generated projective module, and all ﬁnitely generated projective C (X )modules arise
faithfully in this way: the global section functor gives a categorical equivalence
between vector bundles on X and ﬁnitely generated projective modules over C (X ).
This is a celebrated theorem of R.G. Swan, and Section X is devoted to giving
a selfcontained discussion of it, starting from the deﬁnition of a vector bundle.
In particular, via Swan’s Theorem basic results on the tangent bundles of compact manifolds translate into examples of ﬁnitely generated projective modules: for
instance, an Euler characteristic argument shows that the tangent bundle of any
evendimensional sphere S 2k is nontrivial, and thus Γ(T S 2k ) is a ﬁnitely generated
nonfree C (S 2k )module! Following Swan, we will show that examples of nonfree
projective modules over more traditional rings like ﬁnitely generated Ralgebras
follow from examples like these.
Example: Let k be a ﬁeld and R = k [t1 , . . . , tn ] be the polynomial ring over k
in n indeterminates. When n = 1, R is a PID, so indeed every ﬁnitely generated
Rmodule is projective. For n > 1, the situation is much less clear, but the problem
of freeness of ﬁnitely generated projective Rmodules can be stated geometrically as
follows: is any algebraic vector bundle on aﬃne nspace An algebraically trivial?
/k
When k = C, the space AnC = Cn in its usual, Euclidean topology is contractible,
/
which by basic topology implies that any continuous Cvector bundle on An is (continuously) trivial. Moreover, relatively classical complex variable theory shows that
any holomorphic vector bundle on An is (holomorphically) trivial. But asking the
transition functions and the trivialization to be algebraic – i.e., polynomial functions – is a much more stringent problem. In his landmark 1955 paper FAC, J.P.
Serre noted that this natural problem remained open for algebraic vector bundles:
he was able to prove only the weaker result that a ﬁnitely generated projective
Rmodule M is stably free – i.e., there exists a ﬁnitely generated free module
M such that M ⊕ F is free. This became known as Serre’s Conjecture (to
his dismay) and was ﬁnally resolved independently in 1976 by D. Quillen [Qui76]
and A. Suslin [Su76]: indeed, every ﬁnitely generated projective Rmodule is free.
Quillen received the Fields Medal in 1978. Fields Medals are not awarded for the
solution of any single problem, but the prize committee writes an oﬃcial document describing the work of each winner that they found particularly meritorious.
In this case, it was made clear that Quillen’s resolution of Serre’s Conjecture was
one of the reasons he received the prize. All this for modules over a polynomial ring! COMMUTATIVE ALGEBRA 37 For more information on Serre’s Conjecture, the reader could do no better than
to consult a recent book of T.Y. Lam [Lam06].
Exercise (K0 (R)): From a commutative ring R, we will construct another commutative ring K0 (R) whose elements correspond to formal diﬀerences of ﬁnite rank
projective modules. More precisely:
a) Let M0 (R) denote the set of all isomorphism classes of ﬁnitely generated projective modules. For ﬁnitely generated projective modules P and Q we deﬁne
[P ] + [Q] = [P ⊕ Q],
[P ] · [Q] = [P ⊗ Q].
Check that this construction is welldeﬁned on isomorphism classes and endows
M0 (R) with the structure of a commutative semiring with unity. What are the
additive and mulitplicative identity elements?
b) Deﬁne K0 (R) as the Grothendieck group of M0 (R), i.e., as the group completion
of the commutative monoid M0 (R). Convince yourself that K0 (R) has the structure
of a semiring. The elements are of the form [P ] − [Q], and we have [P1 ] − [Q1 ] =
[P2 ] − [Q2 ] ⇐⇒ there exists a ﬁnitely generated projective Rmodule M with
∼
P1 ⊕ Q2 ⊕ M = P2 ⊕ Q1 ⊕ M.
In particular, if P and Q are projective modules, then [P ] = [Q] in K0 (R) iﬀ [P ]
and [Q] are stably isomorphic, i.e., iﬀ they become isomorphic after taking the
direct sum with some other ﬁnitely generated projective module M . c) Show that
we also have [P ] = [Q] iﬀ there exists a nitely generated free module Rn such that
P ⊕ Rn ∼ Q ⊕ Rn . In particular, [P ] = [0] = 0 iﬀ P is stably free: there exists a
=
ﬁnitely generated free module F such that P ⊕ F is free.
d) Show that M0 (R) is cancellative iﬀ every stably free ﬁnitely generated projective
module is free.
e)* Find a ring R admitting a ﬁnitely generated projectve module which is stably
free but not free.
f) Show that the mapping Rn → [Rn ] induces an injective homomorphism of rings
˜
Z → K0 (R). Deﬁne K0 (R) to be the quotient K0 (R)/Z. Show that if R is a PID
˜ 0 (R) = 0.
then K
The following exercise explains why in the deﬁnition of K0 (R) we restrict to ﬁnitely
generated projective modules.
Exercise (Eilenberg Swindle): Let us say that a projective module P is weakly
stably free if there exists a not necessarily ﬁnitely generated free module F such
that P ⊕ F is free. Show that every projective module is weakly stably free. (Hint:
if P ⊕ Q is free, take F = P ⊕ Q ⊕ P ⊕ Q ⊕ . . ..)
3.6. Injective modules.
3.6.1. Basic equivalences.
Although we will have no use for them in the sequel of these notes, in both commutative and (especially) homological algebra there is an important class of modules
“dual” to the projective modules. They are characterized as follows. 38 PETE L. CLARK Proposition 28. For a module E over a ring R, the following are equivalent:
(ii) If ι : M → N is an injective Rmodule homomorphism and φ : M → E is
any homomorphism, there exists at least one extension of φ to a homomorphism
Φ : N → E.
(iii) If ι : M → N is an injection, the natural map Hom(N, E ) → Hom(M, E ) is
injective.
(iv) The (contravariant) functor Hom(E, ) is exact.
(v) Any short exact sequence of Rmodules
ι 0→E→M →N →0
splits: there exists an Rmodule map π : M → E such that π ◦ ι = 1E and thus an
internal direct sum decomposition M = ι(E ) ⊕ ker(π ) ∼ E ⊕ N .
=
A module satisfying these equivalent conditions is called injective.
Exercise: Prove Proposition 28.
Exercise: Show that an Rmodule E is injective iﬀ whenever E is a submodule
of a module M , E is a direct summand of M .
Remark: Note that the set of equivalent conditions starts with (ii)! This is to
facilitate direct comparison to Proposition 18 on projective modules. Indeed, one
should check that each of the properties (ii) through (v) are duals of the corresponding properties for projective modules: i.e., they are obtained by reversing all
arrows. The diﬃculty here with property (i) is that if one literally reverses the
arrows in the deﬁnition of free Rmodule to arrive at a “cofree” Rmodule, one gets
a deﬁnition which is unhelpfully strong: the “cofree Rmodule on a set X ” does
not exist when #X > 1! This can be remedied by giving a more reﬁned deﬁnition
of cofree module. For the sake of curiosity, we will give it later on in the exercises,
but to the best of my knowledge, cofree Rmodules by any deﬁnition do not play
the fundamental role that free Rmodules do.
Exercise X.X: Show that every module over a ﬁeld is injective.
Exercise X.X: Show that Z is not an injective Zmodule. (Thus injectivity is the
ﬁrst important property of modules that is not satisﬁed by free modules.)
Exercise X.X: Let {Mi }i∈I be any family of Rmodules and put M =
Show that M is injective iﬀ Mi is injective for all i ∈ I . ∏
i ∈I Mi . Exercise: For a ring R, show TFAE:
(i) R is absolutely projective: every Rmodule is projective.
(ii) R is absolutely injective: every Rmodule is injective.
3.6.2. Baer’s Criterion.
Theorem 29. (Baer’s Criterion [Bae40]) For a module E over a ring R, TFAE:
(i) E is injective.
(ii) For every ideal nonzero I of R, every Rmodule map φ : I → E extends to an
Rmodule map Φ : R → E . COMMUTATIVE ALGEBRA 39 Proof. (i) =⇒ (ii): this is a special case of condition (ii) of Proposition 28: take
M = I , N = R.
(ii) =⇒ (i): Let M be an Rsubmodule of N and φ : M → E an Rmodule map.
We need to show that φ may be extended to N . Now the set P of pairs (N ′ , φ′ )
with M ⊂ N ′ ⊂ N and φ : N ′ → E a map extending φ is nonempty and has an
evident partial ordering, with respect to which the union of any chain of elements
in P is again an element of P . So by Zorn’s Lemma, there is a maximal element
φ′ : N ′ → E . Our task is to show that N ′ = N .
Assume not, and choose x ∈ N \ N ′ . Put
I = (N ′ : x) = {r ∈ R  rx ⊂ N ′ };
one checks immediately that I is an ideal of R (a generalization to modules of the
colon ideal we have encountered before). Consider the composite map
·x φ I → N ′ → E;
by our hypothesis, this extends to a map ψ : R → E . Now put N ′′ = ⟨N ′ , x⟩ and
deﬁne12 φ′′ : N ′′ → E by
φ′′ (x′ + rx) = φ′ (x′ ) + ψ (r).
Thus φ′′ is an extension of φ′ to a strictly larger submodule of N than N ′ , contradicting maximality.
Exercise X.X: Verify that the map φ′′ is welldeﬁned.
3.6.3. Divisible modules.
Recall that a module M over a domain R is divisible if for all r ∈ R• the endomorphism r• : M → M, x → rx, is surjective. Further, we deﬁne M to be
uniquely divisible if for all r ∈ R• , the endomorphism r• : M → M is a bijection.
Example: The Zmodules Q and Q/Z are divisible. Q is moreover uniquely divisible but Q/Z is not.
Exercise X.X: Show that a divisible module is uniquely divisible iﬀ it is torsionfree.
Exercise X.X: a) Show that a quotient of a divisible module is divisible.
b) Show that arbitrary direct sums and direct products of divisible modules are
divisible.
Exercise X.X: Let R be a domain with fraction ﬁeld K .
a) Show that K is a uniquely divisible Rmodule.
b) Let M be any Rmodule. Show that the natural map M ⊗ M ⊗R K is injective
iﬀ M is torsionfree.
c) Show that for any Rmodule M , M ⊗R K is uniquely divisible.
d) Show that K/R is divisible but not uniquely divisible.
Exercise X.X:
12Since N ′′ need not be the direct sum of N ′ and ⟨x⟩, one does need to check that φ′′ is
welldeﬁned; we ask the reader to do so in an exercise following the proof. 40 PETE L. CLARK a) Show that a Zmodule is uniquely divisible iﬀ it can be endowed with the (compatible) structure of a Qmodule, and if so this Qmodule structure is unique.
b) Show that a Zmodule M is a subgroup of a uniquely divisible divisible Zmodule
iﬀ it is torsionfree.
Proposition 30. Let R be a domain and E an Rmodule.
a) If E is injective, it is divisible.
b) If E is torsionfree and divisible, it is injective.
c) If R is a PID and E is divisible, it is injective.
Proof. a) Let r ∈ R• . Since r is a domain, the map r• : R → R is injective. For
x ∈ E , consider the Rmodule homomorphism φ : R → E given by 1 → x. Since E
is injective, this extends to an Rmodule map φ : R → E . Then Φ(1) ∈ E has the
property that rΦ(1) = Φ(r1) = Φ(r) = φ(1) = x, so r• is surjective on E .
b) Let I be a nonzero ideal of R and φ : I → E be an Rmodule map. For each
a ∈ I • , there is a unique ea ∈ E such that φ(a) = aea . For b ∈ I • , we have
baea = bφ(a) = φ(ba) = aφ(b) = abeb ;
since E is torsionfree we conclude ea = eb = e, say. Thus we may extend φ to a
map Φ : R → E by Φ(r) = re. Thus E is injective by Baer’s Criterion.
c) As above it is enough to show that given a nonzero ideal I of R, every homomorphism φ : I → E extends to a homomorphism R → E . Since R is a PID, we
may write I = xR for x ∈ R• . Then, as in part a), one checks that φ extends to Φ
iﬀ multiplication by x is surjective on M , which it is since M is divisible.
By combining Proposition 30 with Exercise X.X, we are able to show an important
special case of the desired fact that every Rmodule can be realized as a submodule of an injective module. Namely, if M is a torsionfree module over a domain R,
then M is a submodule of the uniquely divisible – hence injective – module M ⊗R K .
Exercise: Let n ∈ Z+ .
a) Show that as a Zmodule Z/nZ is not divisible hence not injective.
b) Show that as a Z/nZ module Z/nZ is divisible iﬀ n is a prime number.
c) Show that Z/nZ is always injective as a Z/nZmodule.
Exercise: Let R = Z[t] and let K be its fraction ﬁeld. Show that the Rmodule
K/R is divisible but not injective.
Exercise: Let R be a domain with fraction ﬁeld K .
a) If R = K , (of course) all Rmodules are both injective and projective.
b) If R ̸= K , the only Rmodule which is both projective and injective is 0.
3.6.4. Enough injectives.
The idea of this section is to pursue the dual version of the statement “Every
Rmodule is a quotient of a projective module”: namely we wish to show that
every Rmodule is a submodule of an injective module. This is a good example
of a statement which remains true upon dualization but becomes more elaborate
to show. The projective version is almost obvious: indeed, we have the stronger
result that every module is a quotient of a free module, and – as we have seen – to
realize M as a quotent of a free Rmodule is equivalent to simply choosing a set of COMMUTATIVE ALGEBRA 41 generators for M . (But again, if we choose the most obvious deﬁnition of “cofree”,
then this statement will be false.)
Let k be a ring, R a k algebra, M an Rmodule and N a k module. Consider
the commutative group Homk (M, N ). We may endow it with the structure of an
Rmodule as follows: for r ∈ R and f ∈ HomZ (M, N ), (rf )(x) := f (rx).
Consider the special case k = Z and N = Q/Z of the above construction. It
gives HomZ (M, Q/Z) the structure of an Rmodule, which we denote by M ∗ and
call the Pontrjagin dual of M .13 Because Q/Z is an injective Zmodule, the (contravariant) functor M → M ∗ – or in other words HomZ ( , Q/Z) – is exact.14 In
particular, if f : M → N is an Rmodule map, then f injective implies f ∗ surjective
and f surjective implies f ∗ injective.
As is often the case for “duals”, we have a natural map M → M ∗∗ : namely
x → (f → f (x)).
Lemma 31. For any Rmodule M , the natural map ΨM : M → M ∗∗ is injective.
Proof. Seeking a contradiction, let x ∈ M • be such that Ψ(x) = 0. Unpacking the
deﬁnition, this means that for all f ∈ HomZ (M, Q/Z), f (x) = 0. But since Q/Z is
an injective Zmodule, it suﬃces to ﬁnd a nontrivial homomorphism Zx → Q/Z,
1
and this is easy: if x has ﬁnite order n > 1, we may map x to n , whereas if x has
inﬁnite order we may map it to any nonzero element of Q/Z.
Lemma 32. Every Zmodule M can be embedded into an injective Zmodule.
⊕
Proof. Let I be a set of generators for the Zmodule M ∗ and put F =
i∈I Z,
and consider the induced surjection of Z modules π : F → M ∗ . Taking ⊕
Pontrjagin
⊕
duals, we get an injection π ∗ : M ∗∗ → F ∗ = Hom( i∈I Z, Q/Z) =
i∈I Q/Z.
Therefore we get an injection
⊕
π ∗ ◦ ΨM : M →
Q/Z.
i ∈I ⊕ The Zmodule i∈I Q/Z is a direct sum of divisible modules, hence divisible, hence
– since Z is a PID – injective. We are done!
Lemma 33. (Injective Production Lemma) Let R be a k algebra, E an injective
k module and F a free Rmodule. Then Homk (F, E ) is an injective Rmodule.
Proof. We will show that the functor HomR ( , Homk (F, E )) is exact. For any Rmodule M , the adjointness of ⊗ and Hom gives
HomR (M, Homk (P, E )) = Homk (F ⊗R M, E )
so we may look at the functor M → Homk (F ⊗R M, E ) instead. This is the
composition of the functor M → F ⊗R M with the functor N → Homk (N, E ).
But we claim that both of these functors are exact. In the former case a moment’s
thought shows this to be true. The latter case is one of our deﬁning properties of
injective modules.
13Recall that the notation M ∨ has already been taken: this is the linear dual Hom (M, R).
R
14Here we are using the (obvious) fact that a sequence of Rmodules is exact iﬀ it is exact when viewed merely as a sequence of Zmodules. 42 PETE L. CLARK Remark: Soon enough we will deﬁne a ﬂat Rmodule to be an Rmodule N such
that the functor M → M ⊗R N is exact. Then Lemma 33 can be rephrased with
the hypothesis that F is a at Rmodule, and (since as we have just seen, free
Rmodules are ﬂat) this gives a somewhat more general result.
Theorem 34. Every Rmodule can be embedded into an injective Rmodule.
Proof. Let M be an Rmodule. Viewing M as a Zmodule, by Lemma 32 there
is an injective Zmodule E1 and a monomorphism of Zmodules φ1 : M → E1 .
Further, by Lemma 33, HomZ (R, E1 ) is an injective Rmodule. Now consider the
Rmodule map
φ : M → HomZ (R, E1 ), x → (r → φ1 (rx)).
We claim that φ is a monomorphism into the injective Rmodule HomZ (R, E1 ).
Indeed, if φ(x) = 0 then for all r ∈ R, φ1 (rx) = 0. In particular φ1 (x) = 0, so since
φ1 is a monomorphism, we conclude x = 0.
Exercise: Let us say that a Zmodule is cofree if it is of the form F ∨ for a free
Zmodule F . Then the proof of Lemma 32 gives the stronger statement that every
Zmodule can be embedded into a cofree Zmodule. Formulate a deﬁnition of
cofree Rmodule so that the proof of Theorem 34 gives the stronger statement
that every Rmodule can be embedded into a cofree Rmodule. (Hint: remember
to pay attention to the diﬀerence between direct sums and direct products.)
3.6.5. Essential extensions and injective envelopes.
The results of this section are all due to B. Eckmann and A. Schopf [ES53].
Proposition 35. Let M be an Rmodule and M ⊂R N an Rsubmodule. TFAE:
(i) If X is any nonzero Rsubmodule of N , then X ∩ M is nonzero.
(ii) If x ∈ N • , there exists r ∈ R such that rx ∈ M • .
(iii) If φ : N → Y is an Rmodule map, then φ is injective iﬀ φM is injective.
An extension M ⊂ N satisfying these equivalent conditions is called essential.
Proof. (i) =⇒ (ii): Apply (i) with X = ⟨x⟩.
(ii) =⇒ (iii): Assuming (ii), let φ : N → Y be a homomorphism with φM is
injective. It is enough to show that φ is injective. Seeking a contradiction, let
x ∈ N • be such that φ(x) = 0. By (ii), there exists r ∈ R such that rx ∈ M • . But
then by assumption rφ(x) = φ(rx) ̸= 0, so φ(x) ̸= 0, contradiction.
(iii) =⇒ (i): We go by contraposition. Suppose there exists a nonzero submodule
X of N such that X ∩ M = 0. Then the map φ : N → N/X is not an injection but
its restriction to M is an injection.
Proposition 36. (Tower Property of Essential Extensions) Let L ⊂ M ⊂ N be
Rmodules. Then L ⊂ N is an essential extension iﬀ L ⊂ M and M ⊂ N are both
essential extensions.
Proof. Suppose ﬁrst that L ⊂ N is an essential extension. Then for any nonzero
submodule X of N , X ∩ L ̸= 0. In particular this holds for X ⊂ M , so L ⊂ M is
essential. Moreover, since L ⊂ M , X ∩ L ̸= 0 implies X ∩ M ̸= 0, so M ⊂ N is
essential. Conversely, suppose L ⊂ M and M ⊂ N are both essential, and let X be
a nonzero submodule of N . Then X ∩ M is a nonzero submodule of M and thus
(X ∩ M ) ∩ L = X ∩ L is a nonzero submodule of L. So L ⊂ N is essential. COMMUTATIVE ALGEBRA 43 So why are we talking about essential extensions when we are supposed to be talking
about injective modules? The following result explains the connection.
Theorem 37. For an Rmodule M , TFAE:
(i) M is injective.
(ii) M has no proper essential extensions: i.e., if M ⊂ N is an essential extension,
then M = N .
N . By Exercise X.X, M is a
Proof. (i) =⇒ (ii): Let M be injective and M
direct summand of N : there exists M ′ such that M ⊕ M ′ = N . Thus M has zero
intersection with M ′ , and by criterion (ii) of Proposition 35, we must have M ′ = 0
and thus M = N .
(ii) =⇒ (i): let N be an Rmodule such that M ⊂ N . To show that M is injective,
by Exercise X.X it is equivalent to show that M is a direct summand of N . Now
consider the family of submodules M ′ of N with the property that M ∩ M ′ = 0.
This family is partially ordered by inclusion, nonempty, and closed under unions
of chains, so by Zorn’s Lemma there exists a maximal such element M ′ . Now
consider the extension M → N/M ′ : we claim it is essential. Indeed, if not, there
exists x ∈ N \ M ′ such that ⟨M ′ , x⟩ ∩ M = 0, contradicting maximality of M ′ .
But by hypothesis, M has no proper essential extensions: thus M = N/M ′ , i.e.,
M ⊕ M ′ = N and M is a direct summand of N .
We say that an extension M ⊂ N is maximal essential if it is essential and
there is no proper extension N ′ of N such that M ⊂ N ′ is essential. Combining
Proposition 36 and Theorem 37 yields the following important result.
Theorem 38. For an essential extension M ⊂ N of Rmodules, TFAE:
(i) M ⊂ N is maximal essential.
(ii) N is injective.
Exercise: To be sure you’re following along, prove Theorem 38.
Once again we have a purpose in life – or at least, this subsubsection of it – we
would like to show that every Rmodule admits a maximal essential extension and
that such extensions are unique up to isomorphism over M . Moreover, a plausible
strategy of proof is the following: let M be an Rmodule. By Theorem 34 there
exists an extension M ⊂ E with E injective. Certainly this extension need not be
essential, but we may seek to construct within it a maximal essential subextension
N and then hope to show that M ⊂ E ′ is injective.
Theorem 39. Let M be an Rmodule and M ⊂ E an extension with E an injective
module. Let P be the set of all essential subextensions N of M ⊂ E . Then:
a) P contains at least one maximal element.
b) Every maximal element E ′ of P is injective.
Proof. The proof of part a) is the usual Zorn’s Lemma argument: what we need
to check is that the union N of any chain {Ni } of essential subextensions is again
an essential subextension. Suppose for a contradiction that there exists a nonzero
submodule X of N such that X ∩ M = 0. Choose x ∈ X • and put X ′ = ⟨x⟩. Then
X ′ ⊂ Ni for some i and X ′ ∩ M ⊂ X ∩ M = 0, contradicting the essentialness (?!)
of the extension M ⊂ Ni .
Now let E ′ be a maximal essential subextension of M ⊂ E . We need to show 44 PETE L. CLARK that M ⊂ E ′ is actually a maximal essential extension: so suppose there is an
essential extension E ′ ⊂ N . Let ι : M ⊂ E ′ ⊂ N be the composite map. It is
a monomorphism, so by the injectivity of E the injection M ⊂ E extends to a
homomorphism φ : N → E . But φM is an injection and M ⊂ N is an essential
extension, so by condition (iii) of Proposition 35 this implies that φ itself is an
injection. By maximality of E ′ among essential subextensions of M ⊂ E we must
have E ′ = N .
For an Rmodule M , we say that an extension M ⊂ E is an injective envelope
(other common name: injective hull) of M if M ⊂ E is a maximal essential
extension; equivalently, an essential extension with E injective. Thus Theorem 39
shows that any Rmodule admits an injective envelope.
Proposition 40. Let R be an integral domain with fraction ﬁeld K . Then R ⊂ K
is an injective envelope of R.
Exercise: Prove Proposition 40. (Suggestion: use the relationship between injective
modules and divisible modules.)
Exercise: More generally, let M be a torsionfree module over a domain R. Show
that M ⊂ M ⊗R K is an injective envelope of M .
Let us touch up our characterization of injective envelopes a bit.
Proposition 41. (Equivalent Properties of an Injective Envelope) For an extension
M ⊂ E of Rmodules, TFAE:
(i) M ⊂ E is a maximal essential extension.
(ii) M ⊂ E is essential and E is injective.
(iii) E is minimal injective over M : there does not exist any proper subextension
M ⊂ E ′ ⊂ E with E ′ injective.
Proof. We have already seen that (i) ⇐⇒ (ii).
(ii) =⇒ (iii): Assume that E is injective and E ′ is an injective subextension of
M ⊂ E . Since E ′ is injective, there exists N ⊂ E such that E ′ ⊕ N = E . Moreover,
M ∩ N ⊂ E ′ ∩ N = 0, so M ∩ N = 0. Since M ⊂ E is essential, we must have
N = 0, i.e., E ′ = E .
(iii) =⇒ (ii): Suppose that M ⊂ E is minimal injective. The proof of Theorem
39 gives us a subextension E ′ of M ⊂ E such that E ′ is injective and M ⊂ E ′ is
essential. Thus by minimality E = E ′ , i.e., M ⊂ E is essential.
Theorem 42. (Uniqueness of Injective Envelopes) Let M be an Rmodule and
let ι1 : M ⊂ E1 , ι2 : M ⊂ E2 be two injective envelopes of M . Then E1 and
E2 are isomorphic as Rmodule extensions of M : i.e., there exists an Rmodule
isomorphism Φ : E1 → E2 such that Φ ◦ ι1 = ι2 .
Proof. Since ι1 : M → E1 is a monomorphism and E2 is injective, the map ι2 :
M → E2 extends to a map Φ : E1 → E2 such that Φ ◦ ι1 = ι2 . Since the restriction
of Φ to the essential submodule M is a monomorphism, so is Φ. The image Φ(E1 )
is an essential subextension of M ⊂ E2 , so by condition (iii) of Proposition 41 we
must have E2 = Φ(E1 ). Thus Φ : E1 → E2 is an isomorphism.
In view of Theorem 42, it is reasonable to speak of “the” injective envelope of M
and denote it by M → E (M ). Reasonable, that is, but not ideal: it is not true that COMMUTATIVE ALGEBRA 45 any two injective envelopes are canonically isomorphic.15 Otherwise put, formation of the injective envelope is not functorial. For more on this in a more general
categorytheoretic context, see [AHRT02].
Exercise: Let M be a submodule of an injective module E . Show that E contains an isomorphic copy of the injective envelope E (M ).
Exercise: If M ⊂ N is an essential extension of modules, then E (M ) = E (N ).
3.7. Flat modules.
Suppose we have a short exact sequence
0 → M ′ → M → M ′′ → 0
of Rmodules. If N is any Rmodule, we can tensor each element of the sequence
with N , getting by functoriality maps
0 → M ′ ⊗ N → M ⊗ N → M ′′ ⊗ → 0.
Unfortunately this new sequence need not be exact. It is easy to see that it is right
exact: that is, the piece of the sequence
M ′ ⊗ N → M ⊗ N → M ′′ ⊗ N → 0
remains exact. This follows because of the canonical “adjunction” ismorphism
Hom(M ⊗ N, P ) = Hom(M, Hom(N, P ))
and the leftexactness of the sequence Hom( , Y ) for all Rmodules Y . However,
tensoring an injection need not give an injection. Indeed, consider the exact sequence
[2] 0 → Z → Z.
If we tensor this with Z/2Z, we get a sequence
[2] 0 → Z/2Z → Z/2Z,
but now the map Z ⊗ Z/2Z → Z ⊗ Z/2Z takes n ⊗ i → (2n ⊗ i) = n ⊗ 2i = 0, so is
not injective.
Deﬁnition: A module M over a ring R is ﬂat if the functor N → N ⊗R M is
exact. This means, equivalently, that if M → M ′ then M ⊗ N → M ′ ⊗ N , or also
that tensoring a short exact sequence with M gives a short exact sequence.
It will probably seem unlikely at ﬁrst, but in fact this is one of the most important
and useful properties of an Rmodule.
So, which Rmodules are ﬂat?
Proposition 43. Let {Mi }i∈I be a family of Rmodules. TFAE:
(i) For all i, Mi is ﬂat. ⊕
(ii) The direct sum M = i Mi is ﬂat.
15The situation here is the same as for “the” splitting ﬁeld of an algebraic ﬁeld extension or
“the” algebraic closure of a ﬁeld. 46 PETE L. CLARK Exercise: Prove Proposition 43.
Proposition 44. Let R be a domain. Then ﬂat Rmodules are torsionfree.
Proof. We will prove the contrapositive. Suppose that 0 ̸= m ∈ R[tors], and let
0 ̸= r ∈ R be such that rm = 0 Since R is a domain, we have a short exact sequence
[r ] 0 → R → R → R/rR → 0
and tensoring it with M gives
[r ] 0 → M → M → M/rM → 0,
but since rm = 0 the ﬁrst map is not injective.
Proposition 45. Projective Rmodules are ﬂat.
Proof. A projective Rmodule is a module P such that there exists P ′ with P ⊕ P ′ ∼
=
F a free module. Therefore, by Proposition 43, it is enough to show that free
modules are ﬂat. By abuse of notation, we will abbreviate the inﬁnite direct sum
of d copies of R as Rd . Since for any Rmodule M we have M ⊗R Rd = M d , it
follows that tensoring a short exact sequence
0 → M ′ → M → M ′′ → 0
with F = Rd just yields
0 → (M ′ )d → (M )d → (M ′′ )d → 0.
This is still exact.
3.8. Nakayama’s Lemma.
3.8.1. Nakayama’s Lemma.
Proposition 46. Let M be a ﬁnitely generated Rmodule, I an ideal of R, and φ
be an Rendomorphism of M such that φ(M ) ⊂ IM . Then φ satisﬁes an equation
of the form
φn + an−1 φn−1 + . . . + a1 φ + a0 = 0,
with ai ∈ I .
Proof. Let x1 , . . . , xn be a set of generators for M as an Rmodule. Since each
∑
φ(xi ) ∈ IM , we may write φ(xi ) = j aij xj , with aij ∈ I . Equivalently, for all i,
n
∑ (δij φ − aij )xj = 0. j =1 By multiplying on the left by the adjoint of the matrix M = (δij φ − aij ), we get
that det(δij φ − aij ) kills each xi , hence is the zero endomorphism of M . Expanding
out the determinant gives the desired polynomial relation satisﬁed by φ.
Exercise X.X: Some refer to Prop. 46 as the CayleyHamilton Theorem. Discuss.
Theorem 47. (Nakayama’s Lemma) Let R be a ring, J an ideal of R, and M a
ﬁnitely generated Rmodule such that JM = M .
a) There exists x ∈ R with x ≡ 1 (mod J ) such that xM = 0.
b) Suppose moreover that J is contained in every maximal ideal of R. Then M = 0. COMMUTATIVE ALGEBRA 47 Proof. Applying Proposition 46 to the identity endomorphism φ: gives a1 , . . . , an ∈
J such that for x := 1 + a1 + . . . + an , xM = 0 and x ≡ 1 (mod J ), proving part
a). If moreover J lies in every maximal ideal m of R, then x ≡ 1 (mod )m for all
maximal ideals m, hence x lies in no maximal ideal of R. Therefore x is a unit and
xM = 0 implies M = 0.
Corollary 48. Let R be a ring, J an ideal of R which is contained in every maximal
ideal of R, M a ﬁnitely generated Rmodule and N a submodule of M such that
JM + N = M . Then M = N .
Proof. We have J (M/N ) = (JM + N )/N = M/N . Applying Nakayama’s Lemma
to the ﬁnitely generated module M/N , we conclude M/N = 0, i.e., N = M .
Corollary 49. Let R be a ring, J an ideal of R which is contained in every maximal
ideal of R, and M a ﬁnitely generated Rmodule. Let x1 , . . . , xn ∈ M be such that
their images in M/JM span M/JM as an R/J module. Then the xi ’s span M .
Proof. Let N = ⟨x1 , . . . , xn ⟩R , and apply Corollary 48.
Corollary 50. Let R be a ring and J an ideal which is contained in every maximal
ideal of R. Let M and N be Rmodules, with N ﬁnitely generated, and let u : M →
N be an Rmodule map. Suppose that the map uJ : M/JM → N/JN is surjective.
Then u is surjective.
Proof. Apply Nakayama’s Lemma to J and N/M .
Recall that an element x in a ring R such that x2 = x is called idempotent.
Similarly, an ideal I of R such that I 2 = I is called idempotent.
Exercise: Let R be a ring and I an ideal of R.
a) Suppose I = (e) for an idempotent element e. Show that I is idempotent.
b) Give an example of a nonidempotent x such that (x) is idempotent.
c) Is every idempotent ideal generated by some idempotent element?
The last part of the preceding exercise is rather diﬃcult. It turns out that the
answer is negative in general: we will see later that counterexamples exist in any
inﬁnite Boolean ring. However under a relatively mild additional hypothesis the
answer is aﬃrmative.
Corollary 51. Let R be any ring and I a ﬁnitely generated idempotent ideal of
R. Then there exists an idempotent e ∈ R such that I = (e). In particular, in a
Noetherian ring every idempotent ideal is generated by a single idempotent element.
Exercise: Prove Corollary 51. (Hint: apply Theorem 47!)
3.8.2. Hopﬁan modules.
A group G is Hopﬁan if every surjective group homomorphism f : G → G is
an isomorphism – equivalently, G is not isomorphic to any of its proper quotients.
This concept has some currency in combinatorial and geometric group theory.
Some basic examples: any ﬁnite group is certainly Hopﬁan. A free group is Hopﬁan
iﬀ it is ﬁnitely generated, and more generally a ﬁnitely generated residually ﬁnite 48 PETE L. CLARK ∏∞
group is Hopﬁan. An obvious example of a nonHopﬁan group is i=1 G for any
nontrivial group G. A more interesting example is the BaumslagSolitar group
B (2, 3) = ⟨x, y  yx2 y −1 = x3 ⟩.
More generally, let C be a concrete category: that is, Ob C is a class of sets and
for all X, Y ∈ Ob C , HomC (X, Y ) ⊂ HomSet (X, Y ), i.e., the morphisms between X
and Y are certain functions from X to Y . We may deﬁne an object X in C to be
Hopﬁan if every surjective endomorphism of X is an isomorphism.
Exercise X.X:
a) Give an example of a concrete category C and a ﬁnite object X (i.e., the underlying set of X is ﬁnite) which is not Hopﬁan.
b) Suppose that C satisﬁes the property that any bijective morphism is an isomorphism. Show that any ﬁnite object is a Hopﬁan object.
c) In the category of Sets, the Hopﬁan objects are precisely the ﬁnite sets.
Remark: Our discussion of “Hopﬁan objects” in categories more general than RMod is not particularly serious or well thought out. So far as I know there is
not a completely agreed upon deﬁnition of a Hopﬁan object, but Martin Brandenburg has suggested (instead) the following: X ∈ C is Hopﬁan if every extremal
epimorphism X → X is an isomorphism.
Theorem 52. Let R be a ring and M a ﬁnitely generated Rmodule. Then M is
a Hopﬁan object in the category of Rmodules.
Proof. ([M, p. 9]) Let f : M → M be a surjective Rmap. We show f is injective.
There is a unique R[t]module structure on M extending the given Rmodule
structure and such that for all m ∈M, tm = f (m). Let I = tR[t]. By hypothesis
IM = M , so by Nakayama’s Lemma there exists P (t) ∈ R[t] such that (1 +
P (t)t)M = 0. Let y ∈ ker f . Then
0 = (1 + P (t)t)y = y + P (t)f (y ) = y + P (t)0 = y.
So f is injective.
Exercise: Show that (Q, +) is a Hopﬁan Zmodule which is not ﬁnitely generated.
Exercise: Do there exist Hopﬁan Zmodules of all cardinalities? (An aﬃrmative
answer was claim in [Bau62], but it was announced in [Bau63] that the construction
is not valid. So far as I know the problem remains open lo these many years later.)
3.8.3. A variant.
The results of this section are taken from [DM71, §I.1].
Proposition 53. (Generalized Nakayama’s Lemma) Let R be a ring, J an ideal
of R and M a ﬁnitely generated Rmodule. TFAE:
(i) J + ann M = R.
(ii) JM = M .
Proof. (i) =⇒ (ii): If J + ann M = R, we may write 1 = x + y with x ∈ J, y ∈
ann M , so that for all m ∈ M , m = 1m = xm + ym = xm. Thus JM = M .
(ii) =⇒ (i): Conversely, suppose M = ⟨m1 , . . . , mn ⟩. For 1 ≤ i ≤ n, put COMMUTATIVE ALGEBRA 49 Mi = ⟨mi , . . . , mn ⟩ and Mn+1 = 0. We claim that for all 1 ≤ i ≤ n + 1 there exists
ai ∈ J with (1 − ai )M ⊂ Mi , and we will prove this by induction on n. We may
take a1 = 0. Having chosen a1 , . . . , ai , we have
(1 − ai )M = (1 − ai )JM = J (1 − ai )M ⊂ Mi ,
so there exist aij ∈ J such that
(1 − ai )mj = n
∑ aij mj , j =i or
(1 − ai − aii )mi ∈ Mi+1 .
Thus
(
)
1 − (2ai + aii − a2 − ai ai i) M = (1 − ai )(1 − ai − aii )M ⊂ (1 − ai − aii )Mi ⊂ Mi+1 ,
i
and we may take
ai+1 = 2ai + aii − a2 − ai ai i.
i
So there is an ∈ J such that 1 − an ∈ ann M , and thus 1 ∈ J + ann M .
Exercise: Deduce part b) of Nakayama’s Lemma from Proposition 53.
Corollary 54. Let M be a ﬁnitely generated Rmodule such that mM = M for all
maximal ideals of R. Then M = 0.
Exercise: Prove Corollary 54.
For an Rmodule M , we deﬁne its trace ideal to be the ideal T (M ) of R generated by all the images f (M ) of Rmodule maps f ∈ R∨ = HomR (M, R).
Theorem 55. Let P be a ﬁnitely generated projective Rmodule. Then R splits as
a direct product of rings:
R = T (P ) ⊕ ann P.
Proof. Step 1: We show that T (P ) and ann P are comaximal ideals of R, i.e.,
T (P ) + ann P = R. By the Dual Basis Lemma (Proposition 20) and the following
exercise, there exist x1 , . . . , xn ∈ P∑
and f1 , . . . , fn ∈ P ∨ = HomR (P, R) forming
n
a dual basis: for all x ∈ P , x =
i=1 fi (x)xi . By its very deﬁnition we have
fi (x) ∈ T (P ) for all i and x, hence T (P )P = P . By the Generalized Nakayama’s
Lemma (Lemma 53) we have T (P ) + ann P = R.
For any a ∈ ann P , f ∈ P ∨ and x ∈ P we have af (x) = f (ax) = f (0) = 0: thus
T (P ) ∩ ann P = 0. By comaximality, T (P ) ∩ ann P = 0 and the sum is direct.
Corollary 56. A nonzero ﬁnitely generated projective module over a connected
ring R (i.e., without idempotents other than 0 and 1) is faithful.
Exercise: Prove Corollary 56. 50 PETE L. CLARK 3.8.4. Applications to modules over local rings.
Lemma 57. Let R be a ring and J an ideal which is contained in every maximal
ideal of R, and let M be a ﬁnitely presented Rmodule. Suppose that:
(i) M/JM is a free R/J module, and
(ii) The canonical map J ⊗R M → JM is injective.
Then M is a free Rmodule.
Proof. We may choose a family {xi }i∈I of elements of M such that the images in
M/JM give a R/J basis. (Since M is ﬁnitely generated over R, M/JM is ﬁnitely
generated over R/J , so the index set I is necessarily ﬁnite.) Consider the ﬁnitely
⊕
generated free Rmodule L = i∈I R, with canonical basis {ei }. Let u : L → M be
the unique Rlinear mapping each ei to xi , and let K = ker(u). Since M is ﬁnitely
presented, by Proposition 14 K is ﬁnitely generated. We have a commutative
diagram with exact rows:
J ⊗K →J ⊗L→J ⊗M →0
0 → K → L → M → 0,
where each vertical map – a : J ⊗ K → K , b : J ⊗ L → L, c : J ⊗ M → M – is the
natural multiplication map. Our hypothesis is that the the map J ⊗R M → JM is
injective, so by the Snake Lemma we get an exact sequence
u 0 → coker(a) → coker(b) → coker(c).
Now observe that coker(b) = (R/J ) ⊗R L and coker(c) = (R/J ) ⊗R M , and by
deﬁnition the mapping u : L → M gives, upon passage to the quotient modulo J , a
mapping from one R/J module basis to another. So u is an isomorphism and thus
coker(a) = 0, i.e., K/JK = 0. By Nakayama’s Lemma we conclude K = 0, i.e., u
gives an isomorphism from the free module L to M , so M is free.
We can now prove the following result, which is one that we will build upon in our
future studies of modules over commutative rings.
Theorem 58. Let R be a ring with a unique maximal ideal m – i.e., a local ring.
For a ﬁnitely presented Rmodule M , TFAE:
(i) M is free.
(ii) M is projective.
(iii) M is ﬂat.
(iv) The natural map m ⊗R M → mM is an injection.
Proof. Each of the implications (i) =⇒ (ii) =⇒ (iii) =⇒ (iv) is immediate.
Assume (iv). Then, since m is maximal, R/m is a ﬁeld, so every R/mmodule is
free. Therefore Lemma 57 applies to complete the proof.
3.9. Ordinal Filtrations and Applications.
3.9.1. The Transﬁnite D´vissage Lemma.
e
Let M be an Rmodule. By an ordinal ﬁltration on M we mean an ordinal
number α and for each i ≤ α a submodule Mi of M satisfying all of the following:
(OF1) M0 = 0, Mα = M .
(OF2) For all i, j ∈ α + 1, i ≤ j =⇒ Mi ∪ Mj .
⊂
(OF3) For all limit ordinals i ≤ α, Mi = j<i Mj . COMMUTATIVE ALGEBRA 51 So for instance, taking α = ω = {1, 2, 3, . . .} the ﬁrst inﬁnite ordinal, we recover
the usual notion of an exhaustive ﬁltration by submodules Mn , with the additional
∪
convention that Mω = n∈ω Mn .
For i < α, we call Mi+1 /Mi the ith successive quotient. If for a class C of
Rmodules each successive quotient lies in C , we say the ﬁltration is of class C .
Deﬁne the associated graded module Gr(M ) = ⊕
i<α Mi+1 /Mi . Lemma 59. (Transﬁnite D´vissage Lemma) Let M be an Rmodule and {Mi }i≤α
e
an ordinal ﬁltration of M .
a) Suppose we make the following hypothesis:
(DS) For all i < α the submodule Mi is a direct summand of Mi+1 . Then
⊕
M ∼ Gr(M ) =
Mi+1 /Mi .
=
i<α b) Hypothesis (DS) holds if each successive quotient Mi+1 /Mi is projective.
c) Hypothesis (DS) holds if each Mi is injective.
Exercise: Prove Lemma 59. (Hint: transﬁnite induction.)
Corollary 60. For an Rmodule M , TFAE:
(i) M is free.
(ii) M admits an ordinal ﬁltration with successive quotients isomorphic to R.
(iii) M admits an ordinal ﬁltration with free successive quotients.
⊕
Proof. (i) =⇒ (ii): If M is free, then M ∼
=
i∈I R. By the WellOrdering
⊕
Principle16, I is in bijection with an ordinal α, so we may write M ∼ i<α R, and
=
⊕
put Mi = j<i R.
(ii) =⇒ (iii) is immediate.
(iii) =⇒ (i) follows from Lemma 59 since free modules are projective.
3.9.2. Hereditary and semihereditary rings.
An Rmodule is hereditary if every submodule is projective. (In particular a
hereditary module is projective, and thus the property of being projective is “inherited” by its subomdules.) We say that a ring R is hereditary if R is a hereditary
Rmodule, or equivalently every ideal of R is projective as an Rmodule.
Exercise:
a) Show that every submodule of a hereditary module is hereditary.
b) Show that the zero module is hereditary.
c) Show that there are nonzero rings R for which the only hereditary Rmodule is
the zero module.
Example: A PID is a hereditary ring. Indeed, any nonzero ideal of a PID R is
isomorphic as an Rmodule to R.
16Recall this is a settheoretic axiom which is equivalent to the Axiom of Choice and also to
Zorn’s Lemma and that our running convention in these notes to freely use these axioms when
necessary. 52 PETE L. CLARK Vista: We will not study Dedekind domains until much later on, but (especially,
numbertheoretically inclined) readers may nevertheless be familiar with them. It
turns out that any Dedekind domain is a hereditary ring and conversely a hereditary
domain is necessarily a Dedekind domain.
⊕
Theorem 61. Let {Mi }i∈I a family of hereditary Rmodules, put M = i∈I Mi ,
and let πi : M → Mi be projection onto the ith factor. Then, for any submodule N
⊕
of M , N ∼ i∈I πi (N ).
=
Proof. By the WellOrdering Principle there exists a bijection from I to some or⊕
dinal α, and without loss of generality we may assume M = i∈α Mi . For j ∈ α+ ,
⊕
put Pj = i<j + Mi , so that {Mj } is an ordinalindexed chain of Rsubmodules of
M with ﬁnal element Pα = M . For each j ∈ α+ , put
Nj = N ∩ Pj ,
so {Nj } is an ordinal ﬁltration on N with Nα = N . Moreover, for all i ∈ α we have
Ni = Ni+1 ∩ Pi and thus
∼
Ni+1 /Ni = Ni /(Ni+1 ∩ Pi ) = (Ni+1 + Pi )/Pi .
Thus Ni+1 /Ni is isomorphic to a submodule of Pi+1 /Pi ∼ Mi . Since each Mi
=
is hereditary, each successive quotient Ni+1 /Ni is projective, and the Transﬁnite
D´vissage Lemma (Lemma 59) applies to show that
e
⊕
N ∼ Gr N =
Nj +1 /Nj
= = ⊕
j<α+ (N ∩ j<α+ ⊕
i<j ++ Mj )/(N ∩ ⊕
i<j + Mj ) ∼
= ⊕ πj (N ). j<α+ Corollary 62. Let {Mi }i∈I be a family of Rmodules. Then M =
hereditary iﬀ Mi is hereditary for all i. ⊕
i ∈I Mi is Proof. Suppose ﬁrst that each Mi is hereditary, let N be a submodule of M , and
⊕
put Ni = πi (N ). By Theorem 61, N ∼ i∈I Ni . Each Ni is a submodule of the
=
hereditary module M hence is projective. Thus N is a direct sum of projective
modules, hence projective.
Conversely, if M is hereditary, then so are all of its submodules, but the canonical
map ιi : Mi → M realizes Mi as a submodule of M .
Corollary 63. For a ring R, TFAE:
(i) R is hereditary.
(ii) Every free Rmodule is hereditary.
(iii) Every projective Rmodule is hereditary.
Proof. (i) =⇒ (ii) is immediate from Corollary 62.
(ii) =⇒ (iii): Suppose that every free Rmodule is hereditary. Then if P is a
projective Rmodule, P is a submodule of a free module, hence a submodule of a
hereditary module, hence itself heeditary.
(iii) =⇒ (i): R is a projective Rmodule. COMMUTATIVE ALGEBRA 53 ⊕
Theorem 64. Let R be a a PID and F =
i∈I R a free Rmodule. Then any
submodule M of F is again free, of rank less than or equal to the rank of F .
⊕
Proof. Let N be a submodule of F . By Theorem 61, N ∼ i∈I σi (N ), where each
=
σi (N ) is an Rsubmodule of R, i.e., an ideal of R. Since R is a PID, either σi (N )
is the zero module or is isomorphic as an Rmodule to R. Done!
Corollary 65. A projective module over a PID is free.
Proof. A projective module is a direct summand of a free module, and in particular
is a submodule of a free module. Now Theorem 64.
We expect that the following result is familiar to the reader as a special case of the
classiﬁcation of (all) ﬁnitely generated modules over a PID, but while we are here
we may as well give a commutative algebraic proof.
Proposition 66. Let R be a PID. A ﬁnitely generated torsionfree Rmodule is free.
Proof. Let M be ﬁnitely generated and torsionfree. Certainly we may assume that
M is nonzero. Let X be a ﬁnite generating set for M with 0 ∈ X . Let S ⊂ X be
/
a maximal Rlinearly independent subset. Since M is torsionfree, S is not empty.
Let N = ⟨S ⟩R be the Rmodule spanned by S . Clearly N is free with basis S , and
we will be done if we can show N = M .
There is an annoying technicality here: we must check that S is ﬁnite. In fact,
let n = #X , and let S be any ﬁnite Rlinearly independent subset and let s = #S .
We claim that s ≤ n. To see this, we tensor with the fraction ﬁeld K of R, getting
ιK
K s ∼ S ⊗R K → M ⊗R K ∼ K m with m ≤ n. We may conclude that s ≤ m ≤ n
=
=
provided we know that ιK is injective. And this holds because K is a ﬂat Rmodule,
which we will prove later on as a special case of the ﬂatness of localization maps.
Until then, the reader must take this part of the proof on faith.
Say S = {x1 , . . . , xk }. Of course we are done already if S = X , so assume that
X \ S is nonempty. For each y ∈ X \ S there exist ry , r1 , . . . , rk ∈ R, not all zero,
such that ry y = r1 x1 + . . . + rk xk . Then ry ̸= 0, since otherwise by the linear
independence of S all the ri would be zero. In other words, we have ∏
shown that
for each y ∈ X \ S there exists ry ∈ R• such that ry y ∈ N . Put r := y∈X \S ry .
Then rX ⊂ N and thus rM ⊂ N . Now consider the Rmodule homomorphism
L : M → M given by multiplication by r: x → rx. We have just established that
L(M ) ⊂ N , so we may regard L as a homomorphism M → N . Moreover, since M
is torsionfree, L is injective, and therefore L realizes M as a submodule of the free
Rmodule N . By Theorem 64 we conclude that M is free.
Exercise X.X: Let R be a ring with the following property: every submodule of a
ﬁnitely generated free Rmodule is free. Show that R is a principal ring (i.e., every
ideal of R is principal).
An Rmodule M is semihereditary if every ﬁnitely generated submodule is projective. Thus a Noetherian semihereditary module is hereditary. A ring R is semihereditary if R is a semihereditary Rmodule, or equivalently every ﬁnitely generated ideal of R is projective as an Rmodule.
Example: A domain R is semihereditary if every ﬁnitely generated ideal is principal. Such domains are called B´zout domains and will be further studied later
e
on. 54 PETE L. CLARK ⊕
Theorem 67. Let {Mi }i∈I a family of semihereditary Rmodules, put M = i∈I Mi ,
and let πi : M → Mi be projection onto the ith factor. Then, for any ﬁnitely gen⊕
erated submodule N of M , N ∼ i∈I πi (N ).
=
Proof. The proof of Theorem 61 goes through verbatim.
Theorem 68. Let R be a B´zout domain – i.e., a domain in which every ﬁnitely
e
generated ideal is principal – and F a free Rmodule. Then any ﬁnitely generated
submodule N of F is free, of rank less than or equal to the rank of F .
Proof. The proof of Theorem 65 adapts immediately, using Theorem 67 in place of
Theorem 61.
Theorem 69. Let R be a B´zout domain – i.e., a domain in which every ﬁnitely
e
generated ideal is principal. Then every ﬁnitely generated torsionfree Rmodule is
principal.
Proof. The argument is the same as that of Proposition 66 (a special case), using
Theorem 68 in place of Theorem 64.
Theorem 70. (F. Albrecht) Let R be a semihereditary ring, F a free Rmodule,
and P a ﬁnitely generated submodule of F .
a) P is isomorphic to a ﬁnite direct sum of ﬁnitely generated ideals of R.
b) In particular, P is a ﬁnitely generated projective module.
c) If R is a domain with fraction ﬁeld K and F is free of ﬁnite rank n, then the
rank of P – i.e., dimK P ⊗R K – is at most n.
Exercise: Use Theorem 67 to prove Theorem 70.
3.9.3. Big modules.
Lemma 71. (Kaplansky) Let R be a ring, and let F be an Rmodule which is a
⊕
direct sum of countably generated submodules: say F =
λ∈Λ Eλ . Then every
direct summand of F is again a direct sum of countably generated submodules.
Proof. We claim that there is an ordinal ﬁltration {Fi }i≤α on F satisfying all of
the following properties. (i) For all i < α, Fi+1 /Fi is countably generated.
(ii) If Mi = Fi ∩ M, Ni = Fi ∩ N , then Fi = Mi ⊕ Ni . ⊕
(iii) For each i there is a subset Λi of Λ such that Fi = λ∈Λi Λi . sufficiency
of claim: If so, then {Mi }i≤α is an ordinal ﬁltration on M . Moreover, for all i,
since Mi ⊂ Mi+1 are both direct summands of F , Mi is a direct summand of Mi+1 .
Therefore the Transﬁnite D´vissage Lemma (Lemma 59) applies to the ﬁltration
e
on M to give
⊕
Mi+1 /Mi .
M ∼ Gr(M ) =
=
i<α Moreover, for all i < α we have
Fi+1 /Fi = (Mi+1 ⊕ Ni+1 )/(Mi ⊕ Ni ) ∼ Mi+1 /Mi ⊕ Ni+1 /Ni ,
=
which shows that each successive quotient Mi+1 /Mi is countably generated. Therefore M is a direct sum of countably generated submodules.
proof of claim: We will construct the ﬁltration by transﬁnite induction. The
base case and the limit ordinal induction step are forced upon us by the deﬁnition
of ordinal ﬁltration: we must have F0 = {0}, and for any limit ordinal β ≤ α,
∪
assuming we have deﬁned Fi for all i < β we must have Fβ = i<β Fi . COMMUTATIVE ALGEBRA 55 So consider the case of a successor ordinal β = β ′ + 1. Let Q1 be any Eλ which
is not contained in Fβ ′ . (Otherwise we have Fβ ′ = F and we may just deﬁne
Fi = F for all β ≤ i ≤ α.) Let x11 , x12 , . . . be a sequence of generators of Q1 , and
decompose x11 into its M  and N components. Let Q2 be the direct sum of the
ﬁnitely many Eλ which are necessary to write both of these components, and let
x21 , x22 , . . . be a sequence of generators for Q2 . Similarly decompose x12 into M
and N components, and let Q3 be the direct sum of the ﬁnitely many Eλ needed
to write out these components, and let x31 , x32 , . . . be a sequence of generators
of Q3 . We continue to carry out this procedure for all xij , proceeding according to a diagonal enumeration of Z+ × Z+ : i.e., x11 , x12 , x21 , x13 , x22 , x31 , . . .. Put
Fβ = ⟨Fβ ′ , {xij }i,j ∈Z+ ⟩R . This works!
For a cardinal number κ, we say that a module is κgenerated if it admits a generating set of cardinality at most κ.
Exercise (Warﬁeld): Let κ be any inﬁnite cardinal. Formulate and prove a version of Lemma 71 in which each instance of “countably generated” is replaced by
κgenerated.
Theorem 72. (Kaplansky) For a ring R, let Pc be the class of countably generated
projective Rmodules. For an Rmodule M , TFAE:
(i) M admits an ordinal ﬁltration of class Pc .
(ii) M is a direct sum of countably generated projective submodules.
(iii) M is projective.
Proof. (i) ⇐⇒ (ii) follows immediately from Lemma 59.
(ii) =⇒ (iii): any direct sum of projective modules is projective.
(iii) =⇒ (ii): If M is projective, let F be a free Rmodule with F = M ⊕ N .
Certainly F is a direct sum of countably generated submodules (indeed, of singly
generated submodules!), so by Lemma 71 M is a direct sum of a family of countably
generated submodules, each of which must be projective.
While pondering the signiﬁcance of this result, one naturally inquires:
Question 2. Is there a ring R and an Rmodule M which is not a direct sum of
countably generated submodules?
Theorem 73. (CohenKaplansky [CK51], Griﬃth) For a ring R, TFAE:
(i) Every Rmodule is a direct sum of cyclic (i.e., singly generated) Rmodules.
(ii) Every Rmodule is a direct sum of ﬁnitely generated submodules.
(iii) R is an Artinian principal ideal ring.
Building on these results as well as work of Faith and Walker [FW], R.B. Warﬁeld
Jr. proved the following striking results.
Theorem 74. (Warﬁeld [Wa]) Let R be a Noetherian ring which is not a principal
Artinian ring. Then for any cardinal κ, there exists a module M with the following
properties:
(i) M is not κgenerated, and
(ii) Any decomposition of M into the direct sum of nonzero submodules has only
ﬁnitely many direct summands. 56 PETE L. CLARK The hypotheses of Theorem 74 apply for instance to the ring Z of integers and
yields, in particular, for any inﬁnite cardinal κ a commutative group M which is
not a direct sum of κgenerated submodules.
Theorem 75. (Warﬁeld [Wa]) For a ring R, TFAE:
(i) Every Rmodule is a direct sum of cyclic submodules.
(ii) There exists a cardinal number κ such that every Rmodule is a direct sum of
κgenerated submodules.
(iii) R is a principal Artinian ring.
It is natural to wonder whether Theorem 74 can be strengthened in the following
way: an Rmodule M is indecomposable if it cannot be expressed as a direct sum
of two nonzero submodules.
Question 3. For which rings R do there exist indecomposable Rmodules of all
inﬁnite cardinalities?
However, Question 3 has turned out to be bound up with sophisticated settheoretic
considerations. Namely, in a 1959 paper [Fu59], L. Fuchs claimed that there exist
indecomposable commutative groups of all inﬁnite cardinalties, thus giving an aﬃrmative answer to Question 3 for the ring R = Z. However, it was later observed (by
A.L.S. Corner) that Fuchs’ argument is valid only for cardinals κ less than the ﬁrst
inaccessible cardinal. Exactly what an inaccessible cardinal is we do not wish to
say, but we mention that the nonexistence of inaccessible cardinals is equiconsistent
with the standard ZFC axioms of set theory (in other words, if the ZFC axioms
are themselves consistent, then ZFC plus the additional axiom that there are no
inaccessible cardinals remains consistent) but that nevertheless set theorists have
reasons to believe in them. See also [Fu74] in which these issues are addressed
and he proves that there is an indecomposable commutative group of any inﬁnite
nonmeasurable cardinality (note: accessible implies nonmeasurable).
Question 4. Is there a ring R and a projective Rmodule M which is not a direct
sum of ﬁnitely generated submodules?
Again the answer is yes. A very elegant example was given by Kaplansky (unpublished, apparently).17 Namely that R be the ring of all realvalued continuous functions on the unit interval [0, 1], and let I be the ideal of functions f : [0, 1] → R which
vanish near zero: i.e., for which there exists ϵ = ϵ(f ) > 0 such that f [0,ϵ(f )] = 0.
Exercise X.X: Show the ideal I deﬁned above gives a projective Rmodule which is
not the direct sum of ﬁnitely generated submodules. (Suggestions: (i) to show that
I is projective, use the Dual Basis Lemma. (ii) A slick proof of the fact that I is
not a direct sum of ﬁnitely generated submodules can be given by Swan’s Theorem
using the contractibility of the unit interval.)
Lemma 76. Let M be a projective module over the local ring R, and let x ∈ M .
There is a direct summand M ′ of M such that M ′ contains F and M ′ is free.
Proof. Let F be a free module with F = M ⊕ N . Choose a basis B = {ui }i∈I of
F with respect to which the element x of M has the minimal possible number of
nonzero coordinates. Write
x = r1 u1 + . . . + rn un , ri ∈ R• .
17Warm thanks to Gjergji Zaimi for bringing this important example to my attention. COMMUTATIVE ALGEBRA 57 ∑
∑n−1
Then for all 1 ≤ i ≤ n, ri ∈
/
j ̸=i Rrj . Indeed, if say rn =
i=1 si ri , then
∑n−1
x = i=1 ri (ui + si un ), contradicting the minimality of the chosen basis.
Now write ui = yi + zi with yi ∈ M, zi ∈ N , so
∑
∑
(2)
x=
ri u i =
ri yi .
i i We may write
(3) yi = n
∑ cij uj + ti , j =1 with ti a linear combination of elements of B \ {u1 , . . . , un }. Substituting (3) into
(2) and projecting onto M gives the relations
ri = n
∑ cji rj , j =1 or equivalently, for all i,
(1 − cii )ri = ∑ cji rj . j ̸=i If for any i and j , then one of the coeﬃcients of rj in the above equation is a unit
of R, then dividing through by it expresses rj as an Rlinear combination of the
other ri ’s, which as above is impossible. Therefore, since R is local, each coeﬃcient
must lie in the maximal ideal of R:
∀i, 1 − cii ∈ m, ∀i ̸= j, cij ∈ m.
It follows that the determinant of the matrix C = (cij ) is congruent to 1 modulo m,
hence invertible: if x ∈ m and 1 + x is not invertible, then 1 + x = y for y ∈ m, so
1 = y − x ∈ m, contradiction. Therefore replacing u1 , . . . , un in B with y1 , . . . , yn
still yields a basis of F . It follows that M ′ = ⟨y1 , . . . , yn ⟩R is a direct summand of
F hence also of M which is a free module containing x.
Theorem 77. (Kaplansky) Let (R, m) be a local ring, and let P be any projective
Rmodule. Then P is free.
Proof. Step 1: Since by Theorem 72 P is a direct sum of countably generated
projective submodules, we may as well assume that P itself is countably generated.
Step 2: Suppose M = ⟨{xn }∞ ⟩R is a countably generated projective module over
n=1
the local ring R. By Lemma 76, M = F1 ⊕ M1 with F1 free containing x1 . Note
that M1 is again projective and is generated by the images {x′ }∞ of the elements
n n=2
xn under the natural projection map M → M1 . So reasoning as above, we may
write M2 = F2 ⊕ M2 with F2 free containing x′ . Continuing in this manner, we get
2
M= ∞
⊕ Fn , n=1 so M is free.
Exercise X.X: Give an example of a (necessarily inﬁnitely generated) module over
a local PID which is ﬂat but not free.
3.10. Tor and Ext. 58 PETE L. CLARK 3.10.1. Co/chain complexes.
Let R be a ring. A chain complex C• of Rmodules is a family {Cn }n∈Z of
Rmodules together with for all n ∈ Z, an Rmodule map dn : Cn → Cn−1 such
that for all n, dn−1 ◦ dn = 0. (It is often the case that Cn = 0 for all n < 0, but
this is not a required part of the deﬁnition.)
An example of a chain complex of Rmodules is any long exact sequence. However, from the perspective of homology theory this is a trivial example in the following precise sense: for any chain complex we may deﬁne its homology modules:
for all n ∈ Z, we put
Hn (C ) = Ker(dn )/ Im(dn+1 ).
Example: Let X be any topological space. For any ring R, we have the singular
chain complex S (X )• : S (X )n = 0 for n < 0, and for n ≥ 0, S (X )n is the free
Rmodule with basis the set of all continuous maps ∆n → X , where ∆n is the
standard ndimensional simplex. A certain carefully deﬁned alternating sum of restrictions to faces of ∆n gives rise to a boundary map dn : S (X )n → S (X )n−1 , and
the indeed the homology groups of this complex are nothing else than the singular
homology groups Hn (X, R) with coeﬃcients in R.
If C• and D• are two chain complexes of Rmodules, a homomorphism η : C• →
D• is given by maps ηn : Cn → Dn for all n rendering the following inﬁnite ladder
commutative:
INSERT ME!.
In this way one has evident notions of a monomorphism and epimorphisms of
chain complexes. In fact the chain complexes of Rmodules form an abelian category
and thus these notions have a general categorical meaning, but it turns out they
are equivalent to the much more concrete naive conditions: η is a monomorphism
iﬀ each ηn is injective and is an epiomorphism iﬀ each ηn is surjective.
In particular it makes sense to consider a short exact sequence of chain complexes:
0 −→ A• −→ B• −→ C• .
Here is the ﬁrst basic theorem of homological algebra.
Theorem 78. Let
f g 0 −→ A• −→ B• −→ C• −→ 0
be a short exact sequence of chain complexes of Rmodules. Then for all n ∈ Z
there is a natural connecting homomorphism ∂ : Hn (C ) → Hn−1 (A) such that
g ∂ f g ∂ f . . . −→ Hn+1 (C ) −→ Hn (A) → Hn (B ) −→ Hn (C ) −→ Hn−1 (A) −→ . . .
is exact.
Proof. No way. See [W, Thm. 1.3.1].
Moreover, the homology modules Hn are functors: if f : C• → D• is a morphism
of chain complexes, there are induced maps on the homology groups
Hn (f ) : Hn (C ) → Hn (D).
Example: Let f : X → Y be a continuous map of topological spaces. Then for any
basic nchain ∆n → X in S (X )n , composition with f gives a basic nchain ∆n → Y COMMUTATIVE ALGEBRA 59 in S (Y )n and thus a homomorphism of chain complexes S (f ) : S (X )• → S (Y )• .
There are induced maps on homology, namely the usual maps
Hn (f ) : Hn (X, R) → Hn (Y, R).
There is an entirely parallel story for cochain complexes of Rmodules, which are
exactly the same as chain complexes but with a diﬀerent indexing convention: a
cochain complex C • consists of for each n ∈ Z+ an Rmodule C n and a “coboundary
map” dn : C n → C n+1 . To any cochain complex we get cohomology modules:
for all n ∈ Z, put
H n (C ) = Ker(dn )/ Im(dn−1 ).
The rest of the discussion proceeds in parallel to that of chain complexes (including
the realization of singular cohomology as a special case of this construction).
3.10.2. Chain homotopies.
Let C• , D• be two chain complexes, and let f, g : C• → D• be two homomorphisms between them. We say that f and g are chain homotopic if there exist
for all n ∈ Z+ Rmodule maps sn : Cn → Dn+1 such that
fn − gn = dn+1 sn + sn−1 dn .
The sequence {sn } is called a chain homotopy from f to g .
Exercise X.X: Show that chain homotopy is an equivalence relation on morphisms
from C• to D• .
What on earth is going on here? Again topology is a good motivating example: we
say that two maps f, g : X → Y are homotopic if there exists a continuous map
F : X × [0, 1] → Y such that for all x ∈ X , F (x, 0) = f (x) and F (x, 1) = g (x).
This is an equivalence relation and is generally denoted by f ∼ g . We then deﬁne
two topological spaces to be homotopy equivalent if there exist maps φ : X → Y
and ψ : Y → X such that
ψ ◦ φ ∼ 1X , φ ◦ ψ ∼ 1Y .
(We say that φ : X → Y is a homotopy equivalence if there exists a map ψ as
above.) E.g. a space is contractible if it is homotopy equivalent to a single point.
One of the basic tenets of algebraic topology is that it aspires to study topological spaces only up to homotopy equivalence. That is, all of the fundamental invariants of spaces should be the same on homotopy equivalent spaces and
homomorphisms between these invariants induced by homotopic maps should be
identical. Especially, if f : X → Y is a homotopy equivalence, the induced maps
Hn (f ) : Hn (X ) → Hn (Y ) should be isomorphisms. In fact, if f, g : X → Y are homotopic, the induced morphisms S (f ), S (g ) : S (X )• → S (Y )• are chain homotopic.
So the following result ensures that the induced maps on homology are equal.
Proposition 79. If f, g : C• → D• are chain homotopic, then for all n ∈ Z,
Hn (f ) = Hn (g ).
Proof. Replacing f and g by f − g and 0, it is enough to assume that there exists
a chain homotopy s from f to the zero map – i.e., for all n fn = dn+1 sn + sn−1 dn 60 PETE L. CLARK – and show that f induces the zero map on homology. So take x ∈ Hn (C ). Then
x is represented by an element of Cn lying in the kernel of dn , so
fn (x) = dn+1 sn x + sn−1 dn x = dn+1 sn x + 0 = dn+1 sn x.
Thus fn (x) lies in the image of dn+1 Dn+1 → Dn so represents 0 ∈ Hn (D).
3.10.3. Resolutions.
Let M be an Rmodule. A left resolution of M is an inﬁnite sequence {Ai }∞
i=0
of Rmodules, for all n ∈ N an Rmodule map An+1 → An and an Rmodule map
A0 → M such that the sequence
. . . −→ An+1 → An −→ . . . −→ A1 −→ A0 −→ M −→ 0
is exact. By abuse of notation, we often speak of “the resolution A• . Dually, a
right resolution of M is an inﬁnite sequence {B i }∞ of Rmodules, for all n ∈ N
i=0
an Rmodule map B n → B n+1 and an Rmodule map M → B 0 such that the
sequence
0 −→ M −→ B 0 −→ B 1 −→ . . . −→ B n −→ B n1 . . .
is exact. We speak of “the resolution B • ”.
A projective resolution of M is a left resolution A• such that each An is projective. A injective resolution of M is a right resolution B • such that each B n is
injective. (Exactly why we are not interested in left injective resolutions and right
projective resolutions will shortly become clear.)
Theorem 80. (Existence of resolutions) Let M be an Rmodule.
a) Since every Rmodule is the quotient of a projective (indeed, of a free) module,
M admits a projective resolution.
b) Since every Rmodule can be embedded in an injective module, M admits an
injective resolution.
Proof. a) Choose a projective module P0 , a surjection ϵ0 : P0 → M , and put
M0 = ker(ϵ0 ). Inductively, given Mn−1 , we choose a projective module Pn , a
surjection ϵn : Pn → Mn−1 , and put Mn = ker(ϵ0 ). As our map dn : Pn → Pn−1
we take the composite
ϵ n
Pn −→ Mn−1 ker(ϵn−1 ) −→ Pn−1 . We claim that the resulting sequence
. . . −→ Pn+1 → Pn −→ . . . −→ P1 −→ P0 −→ M −→ 0
is exact. It is certainly exact at M . If x ∈ P0 and ϵ0 (x) = 0, then x0 ∈ M0 .
Lifting x0 via the surjection ϵ1 to x1 ∈ P1 , we ﬁnd d1 (x1 ) = ϵ1 (x1 ) = x0 , so
ker(ϵ0 ) ⊂ Im(d1 ). Conversely, since d1 factors through ker(ϵ0 ), it is clear that
Im(d1 ) ⊂ ker(ϵ0 ). Exactly the same argument veriﬁes exactness at Pn for each
n > 0, so P• is a projective resolution of M .
b) We leave the proof of this part to the reader as an exercise, with the following
comforting remark: the notion of an injective module is obtained from the notion
of a projective module by “reversing all the arrows”, which is the same relationship
that a left resolution bears to a right resolution. Therefore it should be possible to
prove part b) simply by holding up the proof of part a) to a mirror. (And it is.) COMMUTATIVE ALGEBRA 61 Theorem 81. (Comparison theorem for resolutions)
a) Let P• be a projective resolution of the Rmodule M . Let N be another Rmodule and f ′ : M → N be an Rmodule map. Then for every left resolution A•
of N there exists a homomorphism η from the chain complex P• → M → 0 to the
chain complex A• → N → 0. Moreover η is unique up to chain homotopy.
b) Let E • be an injective resolution of the Rmodule N . Let M be another Rmodule
and f ′ : M → N be an Rmodule map. Then for every right resolution A• of M
there exists a homomorphism η from the chain complex 0 → M → A• to the chain
complex 0 → N → E • . Moreover η is unique up to chain homotopy.
Proof. No way. See [W, Thms. 2.2.6 and 2.3.7].
Exercise: Let F be a covariant additive functor on the category of Rmodules. Let
C• and D• be two chain complexes of Rmodueles and f, g : C• → D• be two
homomorphisms between them.
a) Show that F C• and F D• are chain complexes and there are induced chain
homomorphisms F f, F g : F C• → F D• .
b) Show that if f and g are chain homotopic, so are F f and F g . (Suggestion: Show
that it makes sense to apply F to a chain homotopy s.)
3.10.4. Derived functors.
Let us consider covariant, additive functors F from the category of Rmodules to itself. (Recall that additive means that for any M, N , the induced map Hom(M, N ) →
Hom(F (M ), F (N )) is a homomorphism of commutative groups.)
Exercise: For any additive functor F and any chain complex C• of Rmodules,
F C• is again a chain complex. (Hint: the point here is that an additive functor
takes the zero homomorphism to the zero homomorphism.)
Thus if
0 −→ M1 −→ M2 −→ M3 −→ 0
is a short exact sequence of Rmodules, then
0 −→ F (M1 ) −→ F (M2 ) −→ F (M3 ) −→ 0
is necessarily a complex of modules but not necessarily exact: it may have nonzero
homology.
Example: For any ring R, the functor F (M ) = M ⊕ M is exact. For R = Z
the functor F (M ) = M ⊗ Z/2Z is not exact: for instance it takes the short exact
sequence
·2
0 −→ Z −→ Z −→ Z/2Z −→ 0
to the complex
·2
0 −→ Z/2Z −→ Z/2Z → Z/2Z −→ 0,
but multiplication by 2 on Z/2Z is not an injection.
Although an exact functor is a thing of beauty and usefulness to all, it turns out
that from a homological algebraic point of view, it is the functors which are “half
exact” which are more interesting: they give rise to co/homology theories. 62 PETE L. CLARK An additive functor F is right exact if for any exact sequence of the form
M1 −→ M2 −→ M3 −→ 0,
the induced sequence
F M1 −→ F M2 −→ F M3 −→ 0
is again exact. Note that this much was true for the functor F (M ) = M ⊗ Z/2Z,
at least for the sequence we chose above. In fact this holds for all tensor products.
Proposition 82. For any ring R and any Rmodule N , the functor F (M ) =
M ⊗R N is right exact.
Exercise: Prove Proposition 82.
We have also the dual notion of an additive functor F being left exact: for any
exact sequence of the form
0 → M1 → M2 → M3 ,
the induced sequence
0 → F M1 → F M2 → F M3
is again exact.
We now wish to press our luck a bit by extending this deﬁnition to contravariant functors. Here a little abstraction actually makes me less confused, so I will
pass it along to you: we say that a contravariant functor F from the abelian category C to the abelian category D is left exact (resp. right exact) if the associated
covariant functor F opp : C opp → D is left exact (resp. right exact). Concretely, a
contravariant functor F from Rmodules to Rmodules is left exact if every exact
sequence of the form
M1 → M2 → M3 → 0
is transformed to an exact sequence
0 → F M3 → F M2 → F M1 .
(And similarly for right exact contravariant functors.)
Proposition 83. Let R be a ring and X be an Rmodule.
a) The functor M → Hom(X, M ) is covariant and left exact.
(Recall that it is exact iﬀ X if projective.)
b) The functor M → Hom(M, X ) is contravariant and left exact.
(Recall that it is exact iﬀ X is injective.)
Exercise X.X: Prove Proposition 83.
Let F be a right exact additive functor on the category of Rmodules. We will
deﬁne a sequence {Ln F }n∈N of functors, with L0 F = F , called the left derived
functors of F . The idea here is that the leftderived functors quantify the failure
of F to be exact.
Let M be an Rmodule. We deﬁne all the functors Ln M at once, as follows: COMMUTATIVE ALGEBRA 63 ﬁrst we choose any projective resolution P• → M → 0 of M . Second we take away
the M , getting a complex P• which is exact except at P0 , i.e.,
H0 (P ) = P0 / Im(P1 → P0 ) = P0 / Ker(P0 → M ) = M,
∀n > 0, Hn (P ) = 0.
Third we apply the functor F getting a new complex F P• . And ﬁnally, we take
homology of this new complex, deﬁning
(Ln F )(M ) := Hn (F P• ).
Now there is (exactly?) one thing which is relatively clear at this point.
Proposition 84. We have (L0 F )(M ) = F M .
Proof. Since P1 → P0 → M → 0 is exact and F is right exact, F P1 → F P0 →
F M → 0 is exact, hence
Im(F P1 → F P0 ) = Ker(F P0 → F M ).
Thus
(L0 F )(M ) = H0 (F P• ) = Ker(F P0 → 0)/ Im(F P1 → F P0 )
= F P0 / Ker(F P0 → F M ) = F M. Before saying anything else about the left derived functors Ln F , there is an obvious
point to be addressed: how do we know they are welldeﬁned? On the face of it,
they seem to depend upon the chosen projective resolution P• of M , which is very
far from being unique. To address this point we need to bring in the Comparison
′
Theorem for Resolutions (Theorem 81). Namely, let P• → M → 0 be any other
projective resolution of M . By Theorem 81, there exists a homomorphism of chain
′
complexes η : P• → P• which is unique up to chain homotopy. Interchanging the
′
′
roles of P• and P• , we get a homomorphism η ′ : P• → P• . Moreover, the composition η ′ ◦ η is a homomorphism from P• to itself, so by the uniqueness η ′ ◦ η is chain
homotopic to the identity map on P• . Similarly η ◦ η ′ is chain homotopic to the
′
identity map on P• , so that η is a chain homotopy equivalence. By Exercise X.X,
′
F η : F P• → F P• is a chain homotopy equivalence, and therefore the induced maps
′
on homology Hn (F η ) : Hn (F P• ) → Hn (F P• ) are isomorphisms. Thus we have
shown that two diﬀerent choices of projective resolutions for M lead to canonically
isomorphic modules (Ln F )(M ).
Exercise: Suppose M is projective. Show that for any right exact functor F and
all n > 0, (Ln F )(M ) = 0.
The next important result shows that a short exact sequence of Rmodules induces
a long exact sequence involving the leftderived functors and certain connecting
homomorphisms (which we have not deﬁned and will not deﬁne here).
Theorem 85. Let
(4) 0 −→ M1 −→ M2 −→ M3 −→ 0 64 PETE L. CLARK be a short exact sequence of Rmodules, and let F be any left exact functor on the
category of Rmodules. Then:
a) There is a long exact sequence
(5)
∂ ∂ . . . → (L2 F )(M3 ) → (L1 F )(M1 ) → (L1 F )(M2 ) → (L1 F )(M3 ) → F M1 → F M2 → F M3 → 0.
b) The above construction is functorial in the following sense: if 0 −→ N1 −→
N2 −→ N3 −→ 0 is another short exact sequence of Rmodules and we have maps
Mi → Ni making a “short commutative ladder”, then there is an induced “long
commutative latter” with top row the long exact sequence associated to the ﬁrst
short exact sequence and the bottom row the long exact sequence associated to the
second short exact sequence.
Proof. No way. See [W, Thm. 2.4.6].
Remark: One says that (5) is the long exact homology sequence associated to
the short exact sequence (4).
Now, dually, if F is a right exact functor on the category of Rmodules, we may
deﬁne right derived functors Rn F . Namely, for an Rmodule M , ﬁrst choose
an injective resolution 0 → M → E • , then take M away to get a cochain complex E • , then apply F to get a cochain complex F E • , and then ﬁnally deﬁne
(Rn F )(M ) = H n (F E • ). In this case, a short exact sequence of modules (4) induces a long exact cohomology sequence
(6)
∂ ∂ 0 → F M1 → F M2 → F M3 → (R1 F )(M1 ) → (R1 F )(M2 ) → (R1 F )(M3 ) → (R2 F )(M1 ) . . .
Exercise: Suppose M is injective. Show that for any left exact functor F and all
n > 0, (Rn F )(M ) = 0.
3.10.5. Tor.
Let M, N be Rmodules, and let F : N → M ⊗R N be the functor “tensor with
M ”. By X.X F is right exact so has left derived functors (Ln F ). By deﬁnition, for
all n ∈ N,
Torn (M, N ) := (Ln F )(N ).
Now un/fortunately the situation is even a little richer than the general case of
leftderived functors discussed above. Namely, the tensor product is really a bifunctor: i.e., a functor in M as well as in N , additive and covariant in each variable
separately. So suppose we took the rightderived functors of M → M ⊗R N and
applied them to M : this would give us Torn (N, M ). So it is natural to ask: how
does Torn (M, N ) compare to Torn (N, M )? Since for n = 0 we have that M ⊗R N
is canonically isomorphic to N ⊗R M , it is natural to hope that the Tor functors
are symmetric. And indeed this turns out to be the case.
Theorem 86. (Balancing Tor) For any Rmodules M and N and all n ≥ 0, there
are natural isomorphisms Torn (M, N ) = Torn (N, M ).
Proof. No way. See [W, Thm. 2.7.2].
Exercise: In order to use the Universal Coeﬃcient Theorem (for homology) in algebraic topology, it is necessary to know the values of Tor1 (M, N ) for any two ﬁnitely COMMUTATIVE ALGEBRA 65 generated Zmodules M and N .
a) Show that for any m, n ∈ Z+ , Tor1 (Z/mZ, Z/nZ) ∼ Z/ gcd(m, n)Z.
=
b) Show that for all Zmodules N , Tor1 (Z, N ) = 0.
c) Explain how the structure theorem for ﬁnitely generated Zmodules reduces the
problem of computation of Tor1 (M, N ) for any ﬁnitely generated M and N to the
two special cases done in parts a) and b).
Exercise: Show that the tor functors commute with direct limits: for all n ∈ N,
any directed system {Mi }i∈I of Rmodules M and any Rmodule N we have a
canonical isomorphism
Torn (lim Mi , N ) → lim Torn (Mi , N ).
−
→
−
→
i i (Suggestion: the case n = 0 is Proposition 17. Use this to show the general case
by brute force: i.e., take a projective resolution of N and track these isomorphisms
through the deﬁnition of Torn .)
3.10.6. Ext.
Nota Bene: At the present time, we do not use the Ext functors for anything
in these notes. However they certainly do appear in commutative algebra and elsewhere. Moreover, having taken the trouble (and it was some trouble!) to set up
enough machinery to deﬁne the Tor functors, we might as well follow it up with the
parallel disucssion of the Ext functors.
Let M, N be Rmodules, and let F : N → Hom(M, N ). By Proposition X.X,
F is covariant and left exact. By deﬁnition, for all n ∈ N,
Extn (M, N ) = (Rn F )(N ).
But again, we have an embarrassment of riches: why didn’t we deﬁne the Ext
functors as the rightderived functors of the contravariant left exact functor G :
N → Hom(N, M )? Again, we can do this.
Theorem 87. (Balancing Ext) Let M and N be Rmodules. Deﬁne functors FM :
N → Hom(M, N ) and GN : M → Hom(M, N ). Then for all n ≥ 0,
(Rn FM )(N ) = (Rn GN )(M ).
Proof. No way. See [W, Thm. 2.7.6].
Exercise: In order to use the Universal Coeﬃcient Theorem (for cohomology) in
algebraic topology, it is necessary to know the values of Ext1 (M, N ) for any two
ﬁnitely generated Zmodules M and N . Compute them. (Hint: as for the analgous
problem with Tor, it is enough to consider four special cases: (M, N ) = (Z, Z),
(M, N ) = (Z, Z/nZ), (M, N ) = (Z/mZ, Z), (M, N ) = (Z/mZ, Z/nZ).)
3.11. More on ﬂat modules.
Theorem 88. (Tensorial Criterion for Flatness) For an Rmodule M , TFAE:
(i) M is ﬂat.
(ii) For every ﬁnitely generated ideal I of R the canonical map I ⊗R M → IM is
an isomorphism. 66 PETE L. CLARK Proof. Fist note that the canonical map I ⊗R M → IM is always a surjection.
(i) =⇒ (ii): if M is ﬂat, then since I → R, I ⊗R M → R ⊗R M = M , so
∼
I ⊗R M → IM .
(ii) =⇒ (i): Every ideal of R is the direct limit of its ﬁnitely generated subideals,
so it follows from Proposition 17 and the exactness of direct limits that I ⊗ M → M
is injective for all ideals I . Moreover, if N is an Rmodule and N ′ ⊂ N is an Rsubmodule, then since N is the direct limit of submodule N ′ + F with F ﬁnitely
generated, to show that N ′ ⊗ M → N ⊗ M is injective we may assume
N = N ′ + ⟨ω1 , . . . , ωn ⟩R .
We now proceed by d´vissage: putting Ni = N ′ + ⟨ω1 , . . . , ωi ⟩R , it is enough to
e
show injectivity at each step of the chain
N ′ ⊗ M → N1 ⊗ M → . . . → N ⊗ M,
and further simplifying, it is enough to show that if N = N ′ + Rω , then N ′ ⊗ M →
N ⊗ M . Let I be the “conductor ideal of N/N ′ ”, i.e., I = {x ∈ R  xω ∈ N ′ }, so
that we get a short exact sequence of Rmodules
0 → N ′ → N → R/I → 0
which gives rise to a long exact homology sequence
. . . → TorR (M, R/I ) → N ′ ⊗ M → N ⊗ M → M/IM → 0.
1
Thus it suﬃces to prove TorR (M, R/I ) = 0. For this we consider the homology
1
sequence associated to
0 → I → R → R/I → 0,
namely
. . . → TorR (M, R) = 0 → TorR (M/R/I ) → I ⊗ M → M → . . . ,
1
1
and from the injectivity of I ⊗ M → M we deduce TorR (M, R/I ) = 0.
1
Theorem 89. (Homological Criterion for Flatness) For a module M over a ring
R, TFAE:
(i) M is ﬂat.
(ii) For every Rmodule N all i > 0, TorR (M, N ) = 0.
i
(ii′ ) For every Rmodule N , TorR (M, N ) = 0.
1
(iii) For every ﬁnitely generated ideal I of R, TorR (M, R/I ) = 0.
1
Proof. (i) =⇒ (ii): This is a statement about projective resolutions, but given
that it is just about the most basic possible one. Namely, let L• → N → 0 be a
projective resolution of N . Then
. . . → Ln ⊗ M → Ln−1 ⊗ M → . . . → L0 ⊗ M
is exact, so TorR (M, N ) = 0 for all i > 0.
i
(ii) =⇒ (ii′ ) and (ii′ ) =⇒ (iii) are both immediate.
(iii) =⇒ (i): For each ﬁnitely generated ideal I of R, the short exat sequence
0 → I → R → R/I → 0
of Rmodules induces a long exact sequence in homology, which ends
. . . → TorR (M, R/I ) = 0 → I ⊗ M → M → M/IM → 0,
1 COMMUTATIVE ALGEBRA 67
∼ i.e., the map I ⊗ M → M is injective and thus induces an isomorphism I ⊗ M → IM .
Using the Tensorial Criterion for Flatness (Theorem 88), we conclude M is ﬂat.
Corollary 90. (Direct limits preserve ﬂatness) Let R be a ring and {Mi }i∈I a
directed system of ﬂat Rmodules. Then M = lim Mi is a ﬂat Rmodule.
−
→
Proof. For every Rmodule N , we have
TorR (lim Mi , N ) ∼ TorR (N, lim Mi ) = lim TorR (N, Mi ) ∼ lim TorR (Mi , N ) = lim 0 = 0.
=
=−
1−
1
1
1
→
−
→
−
→
→
−
→
Now apply the Homological Criterion for Flatness.
Corollary 91. For a domain R, TFAE:
(i) Every ﬁnitely generated torsionfree Rmodule is ﬂat.
(ii) Every torsionfree Rmodule is ﬂat.
Proof. Every submodule of a torsionfree Rmodule is torsionfree, and every Rmodule is the direct limit of its ﬁnitely generated submodules. So the result follows
immediately from Proposition 90.
Corollary 92. Let 0 → M ′ → M → M ′′ → 0 be a short exact sequence of Rmodules, with M ′′ ﬂat. Then M ′ is ﬂat iﬀ M is ﬂat.
Exercise X.X: Use the Homological Criterion of Flatness to prove Corollary 92.
Exercise X.X: In a short exact sequence of Rmodules as in Corollary 92, if M ′
and M are ﬂat, must M ′′ be ﬂat?
Now recall that a ﬁnitely generated torsion free module over a PID is free (Proposition 66). From this we deduce:
Corollary 93. A module over a PID is ﬂat iﬀ it is torsionfree.
Exercise: Let R be a domain and M a torsion Rmodule. Show that for all Rmodules N and all n ≥ 0, Torn (M, N ) is a torsion Rmodule.
Theorem 94. Let R be a PID and let M, N be Rmodules.
a) For all n ≥ 2, Torn (M, N ) = 0.
b) Tor1 (M, N ) is a torsion Rmodule.
Proof. a) Choose a free module F0 and a surjection d0 : F0 → N . By X.X, F1 =
Ker(d0 ) is free, so we get a ﬁnite free resolution of N:
0 → F1 → F0 → N → 0.
Therefore we certainly have Torn (M, N ) for all M and all n ≥ 2.
b) Let {Mi }i∈I be the direct system of all ﬁnitely generated submodules of M . As
above, we have M = lim Mi , so
−
→
Tor1 (M, N ) = Tor1 (lim Mi , N )) = lim Tor1 (Mi , N ).
−
→
−
→
By Corollary 93, each Mi which is torsionfree is ﬂat, hence Tor1 (Mi , N ) = 0. Thus
the only possible contribution to lim Tor1 (Mi , N ) comes from torsion modules Mi ,
−
→
and by Exercise X.X, Mi torsion implies Tor1 (Mi , N ) torsion. Thus Tor1 (M, N ) is
a direct limit of torsion modules, hence itself a torsion module. 68 PETE L. CLARK Theorem 95. (Equational Criterion for Flatness) Let M be an Rmodule.
a) Suppose M is ﬂat, and that we are given r, n ∈ Z+ , a matrix A = (aij ) ∈
Mr×n (R) and elements x1 , ldots, xn ∈ M such that
∑
∀1 ≤ i ≤ r,
aij xj = 0.
j Then there exists s ∈ Z+ , bjk ∈ R and yk ∈ M (for 1 ≤ j ≤ n and 1 ≤ k ≤ s) such
that
∑
∀i, k,
aij bj = 0
j and
∀j, xj = ∑ bjk yk = 0. j Thus the solutions in a ﬂat module of a system of linear equations with Rcoeﬃcients
can be expressed as a linear combination of solutions of the system in R.
b) Conversely, if the above conditions hold for a single equation (i.e., with r = 1),
then M is a ﬂat Rmodule.
Proof. a) Let φ : Rn → Rr be the linear map corresponding to multiplication by
the matrix A and let φM : M n → M r be the same for M , so that φM = φ ⊗ 1M .
Let K = Ker φ. Since M is ﬂat, tensoring with M preserves exact sequences, thus
the sequence
ι⊗1 φ K ⊗R M → M n → M r
is exact. By our hypothesis we have φM (x1 , . . . , xn ) = 0, so that we may write
)
(s
∑
βk ⊗ yk
(x1 , . . . , xn ) = (ι ⊗ 1)
k=1 with βk ∈ K and yk ∈ M . Writing out each βk as an element (b1k , . . . , bnk ) ∈ Rn
gives the desired conclusion.
b) We will use the Tensorial Criterion for Flatness to show that M is ﬂat. Let
I = ⟨a1 , . . . , an ⟩ be a ∑
ﬁnitely generated ideal of R. We may write an arbitrary
∑n
n
element z of I ⊗ M as i=1 ai ⊗ mi with mi ∈ M . Let z = i=1 ai mi denote the
image of z in IM ⊂ M . We want to show that z = 0 implies z = 0, so suppose
∑
that i ai mi = 0. By hypothesis, there exist bij ∈ R and yj ∈ M such that for all
∑
∑
j , i ai bij = 0 and for all i, mi = j bij yj . Thus
(
)
∑
∑∑
∑∑
∑
z=
ai ⊗ mi =
ai bij ⊗ yj =
ai bij ⊗ yj =
0 ⊗ yj = 0.
i i j j i j As an application, we can now improve Theorem 58 by weakening the hypothesis
of “ﬁnite presentation” to the simpler one of “ﬁnite generation”.
Theorem 96. Let M be a ﬁnitely generated ﬂat module over the local ring (R, m).
Then for all n ∈ Z+ , x1 , . . . , xn are elements of M such that the images in R/m
are R/mlinearly independent, then x1 , . . . , xn are Rlinearly independent. COMMUTATIVE ALGEBRA 69 Proof. We go by induction on n. Suppose ﬁrst that n = 1, in which case it is
suﬃcient to show that a1 ∈ R, a1 x1 ̸= 0 implies a1 = 0. By the Equational
Criterion for Flatness, there exist b1 , . . . , bs ∈ R such that abi = 0 for all i and
∑
x1 ∈ i bi M . By assumption, x1 does not lie in mM , so that for some i we must
have bi ∈ R× , and then abi = 0 implies a = 0.
Now suppose n > 1, and let a1 , . . . , an ∈ R are such that a1 x1 + . . . + an xn = 0.
Again using the Equational Criterion for Flatness, there are bij ∈ R and y1 , . . . , ys ∈
∑
∑
M such that for all j ,
ai bij = 0 and xi =
bij yj . Since the set of generators
is minimal, by Nakayama’s Lemma their images in M/mM must be R/mlinearly
independent. In particular xn ∈ mM , so that at least one bnj is a unit. It follows
/
∑n−1
that there exist c1 , . . . , cn−1 ∈ R such that an = i=1 ai ci . Therefore
a1 (x1 + c1 xn ) + . . . + an−1 (xn−1 + cn−1 xn ) = 0.
The images in M/mM of the n − 1 elements x1 + c1 xn , . . . , xn−1 + cn−1 xn are
R/mlinearly independent, so by induction a1 = . . . = an−1 = 0. Thus an = 0.
Theorem 97. For a ﬁnitely generated module M over a local ring R, TFAE:
(i) M is free.
(ii) M is projective.
(iii) M is ﬂat.
Proof. For any module over any ring we have (i) =⇒ (ii) =⇒ (iii). So suppose that
M is a ﬁnitely generated ﬂat module over the local ring (R, m). Let (x1 , . . . , xn ) be
a set of Rmodule generators for M of minimal cardinality. By Nakayama’s Lemma
the images of x1 , . . . , xn in R/m are R/mlinearly independent, and then Theorem
96 implies that x1 , . . . , xn is a basis for M as an Rmodule.
A ring R is called absolutely ﬂat if every Rmodule is ﬂat.
Exercise: Show that any quotient of an absolutely ﬂat ring is absolutely ﬂat.
Proposition 98. For a ring R, TFAE:
(i) R is absolutely ﬂat.
(ii) For every principal ideal I of R, I 2 = I .
(iii) Every ﬁnitely generated ideal of R is a direct summand of R.
Proof. (i) =⇒ (ii): Assume R is absolutely ﬂat, and let I = (x) be a principal
ideal. Tensoring the natural inclusion (x) → R with R/(x), we get a an injection
(x) ⊗R R/(x) → R/(x). But this map sends x ⊗ r → xr + (x) = (x), so it is
identically zero. Therefore its injectivity implies that 0 = (x) ⊗R R/(x) ∼ (x)/(x2 ),
=
so (x) = (x2 ).
(ii) =⇒ (iii): Let x ∈ R. Then x = ax2 for some a ∈ R, so putting e = ax we
have e2 = a2 x2 = a(ax2 ) = ax = e, so e is idempotent, and (e) = (x). In general,
for any two idempotents e, f , we have ⟨e, f ⟩ = (e + f − ef ). Hence every ﬁnitely
generated ideal is principal, generated by an idempotent element, and thus a direct
summand.
(iii) =⇒ (i): Let M be an Rmodule, and let I be any ﬁnitely generated ideal
of R. By assumption, we may choose J such that R = I ⊕ J . Therefore J is
projective, so Tor1 (R/I, M ) = Tor1 (J, M ) = 0. By the Homological Criterion for
Flatness, M is ﬂat. 70 PETE L. CLARK The following striking result came relatively late in the game: it is due independently to Govorov [Gov65] and Lazard [Laz64].
Theorem 99. (GovorovLazard) For a module M over a ring R, TFAE:
(i) M is ﬂat.
(ii) There exists a directed family {Fi }i∈I of ﬁnitely generated free submodules of
M such that M = lim Fi .
−
→
Proof. (i) =⇒ (ii): Suppose M = lim Fi is a direct limit of free modules. Then in
−
→
particular M is a direct limit of ﬂat modules, so by Corollary 90 M is ﬂat.
(ii) =⇒ (i): see [Eis, Thm. A6.6].
3.11.1. Flat Base Change.
Proposition 100. (Stability of ﬂatness under base change) Let M be a ﬂat Rmodule, and f : R → S a ring homomorphism. Then S ⊗R M is a ﬂat S module.
Exercise X.X: Prove Proposition 100.
Exercise X.X: Show that the tensor product of ﬂat Rmodules is a ﬂat Rmodule.
Exercise X.X: Let R be a nonzero commutative ring, and n, m ∈ N.
a) Show that Rm ∼ Rn iﬀ m = n.
=
b) Suppose that φ : Rm → Rn is a surjective Rmodule map. Show that m ≥ n.
c)18 Suppose that φ : Rm → Rn is an injective Rmodule map. Show that m ≤ n.
d) Find a noncommutative ring R for which part a) fails.
Theorem 101. (Hom commutes with ﬂat base change) Let S be a ﬂat Ralgebra
and M, N Rmodules with M ﬁnitely presented. Then the canonical map
ΦM : S ⊗R HomR (M, N ) → HomS (M ⊗R S, N ⊗R S )
induced by (s, f ) → (m ⊗ t) → f (m) ⊗ st is an isomorphism.
Proof. (Hochster) It is immediate that ΦR is an isomorphism and that ΦM1 ⊕M2 =
ΦM1 ⊕ ΦM2 , and thus ΦM is an isomorphism when M is ﬁnitely generated free. For
ﬁnitely presented M , there is an exact sequence
H→G→M →0
with H and G ﬁnitely generated free modules. Now we have the following commutative diagram:
0 −→ 0
θM
S ⊗R HomR (M, N ) −→ HomS (M ⊗R S, N ⊗R S )
θ G
S ⊗R HomR (G, N ) −→ HomS (G ⊗R S, N ⊗R S ) θ H
S ⊗R HomR (H, N ) −→ HomS (H ⊗R S, N ⊗R S ).
Note that the right column is obtained by ﬁrst applying the exact functor A →
A ⊗R S and then applying the right exact cofunctor U → HomS (U, N ⊗R S ), so
it is exact. Similarly, the left column is obtained by ﬁrst applying the right exact
cofunctor A → HomR (A, N ) and then applying the exact (since R is ﬂat) functor
A → A ⊗R S , so is exact. Since G and H are ﬁnitely generated free, θG and θH are
isomorphisms, and a diagram chase shows that θM is an isomorphism. 18This is actually quite challenging. COMMUTATIVE ALGEBRA 71 3.12. Faithful ﬂatness.
Proposition 102. For an Rmodule M , TFAE:
(i) For a sequence
(7) α β N1 −→ N2 −→ N3 of left Rmodules to be exact it is necessary and suﬃcient that
(8) A B M ⊗R N1 −→ M ⊗R N2 −→ M ⊗R N3 be exact.
(ii) M is ﬂat and for all nonzero Rmodules N , M ⊗R N ̸= 0.
(iii) M is ﬂat and for all nonzero Rmodule maps u : N → N ′ ,
1M ⊗ u : M ⊗R N → M ⊗R N ′ is not zero .
(iv) M is ﬂat and for every m ∈ MaxSpec R, mM M .
(v) M is ﬂat and for every p ∈ Spec R, pM M .
A module satisfying these equivalent conditions is faithfully ﬂat.
Proof. (i) =⇒ (ii): Certainly (i) implies that M is ﬂat. Moreover, if N is a
nonzero Rmodule such that M ⊗ N = 0, then 0 → N → 0 is not exact but its
tensor product with M is exact, contradicting (i).
(ii) =⇒ (iii): Let I = Im(u); then M ⊗ I = Im(1M ⊗ u). So assuming (ii) and
that I ̸= 0, we conclude Im(1M ⊗ u) ̸= 0.
(iii) =⇒ (i): Assume (iii). Then, since M is ﬂat, if (7) is exact, so is (8).
Conversely, suppose (8) is exact, and put I = Im(α), K = ker(β ). Then B ◦ A =
1M ⊗ (β ◦ α) = 0, so β ◦ α = 0, or in other words, I ⊂ K . We may therefore form
the exact sequence
0 → I → K → K/I → 0,
and tensoring with the ﬂat module M gives an exact sequence
0 → M ⊗ I → M ⊗ K → M ⊗ K/I → 0.
But M ⊗ K = M ⊗ I by hypothesis, so K/I = 0 and I = K .
(ii) =⇒ (iv): Let m ∈ MaxSpec R. Then R/m is a nonzero Rmodule, so by (ii)
so is M ⊗ R/m = M/mM , i.e., mM M .
(iv) =⇒ (ii): Assume (iv) holds. Then, since every proper ideal is contained in
a maximal ideal, we have moreover that for all proper ideals I of R, IM
M , or
equivalently M ⊗ (R/I ) ̸= 0. But the modules of the form R/I as I ranges over
all proper ideals of R are precisely all the cyclic (a.k.a. monogenic) Rmodules, up
to isomorphism. Now if N is any nonzero Rmodule, choose 0 ̸= x ∈ M and let
N ′ = ⟨x⟩ by the cyclic submodule spanned by x. It follows that M ⊗ N ′ ̸= 0. Since
M is ﬂat, N ′ → N implies M ⊗ N ′ → M ⊗ N , so M ⊗ N ̸= 0.
(iv) ⇐⇒ (v): this follows immediately from the proofs of the last two implications,
as we leave it to the reader to check.
Exercise: Show that (iv) ⇐⇒ (v) in Proposition 102.
Corollary 103. Let M be a faithfully ﬂat and u : N → N ′ an Rmodule map.
Then:
a) u is injective iﬀ 1M ⊗ u : M ⊗ N → M ′ N is injective.
b) u is surjective iﬀ 1M ⊗ u is surjective.
c) u is an isomorphism iﬀ 1M ⊗ u is an isomorphism. 72 PETE L. CLARK Exercise: Deduce Corollary 103 from Proposition 102.
Exercise X.X: Use each of the criteria of Proposition 102 to show that the (ﬂat)
Zmodule Q is not faithfully ﬂat.
Exercise X.X: Show that a faithfully ﬂat module is faithful and ﬂat, and that –
unfortunately! – a ﬂat, faithful module need not be faithfully ﬂat.
Exercise X.X: Show that a nonzero free module is faithfully ﬂat but that a nonzero
(even ﬁnitely generated) projective module need not be.
⊕
Exercise X.X: Let {Mi }i∈I be a family of ﬂat Rmodules, and put M = i∈I Mi .
a) Suppose that for some i, Mi is faithfully ﬂat. Show that M is faithfully ﬂat.
b) Give an example where no Mi is faithfully ﬂat yet M is faithfully ﬂat.
Proposition 104. Let 0 → M ′ → M → M ′′ → 0 be a short exact sequence of
Rmodules. Suppose M ′ and M ′′ are ﬂat and that at least one is faithfully ﬂat.
Then M is faithfully ﬂat.
Proof. By Proposition 92, M is ﬂat. Now let N be any Rmodule. Since M ′′ is
ﬂat, Tor1 (M ′′ , N ) = 0 so
0 → M ′ ⊗ N → M ⊗ N → M ′′ ⊗ N → 0
is exact. Thus if M ⊗ N = 0 then M ′ ⊗ N = M ′′ ⊗ N = 0. Since one of M ′ , M ′′
is faithfully ﬂat, by criterion (ii) of Proposition 102 we have N = 0, and then that
same criterion shows that M is faithfully ﬂat.
By a faithfully ﬂat Ralgebra, we mean a ring S equipped with a ring homomorphism R → S making S into a faithfully ﬂat Rmodule.
Proposition 105. Let f : R → S be a ring map and M an Rmodule. Then:
M is faithfully ﬂat iﬀ M ⊗R S is faithfully ﬂat.
Proof. The key fact is that for any S module N , we have
(M ⊗R S ) ⊗S N ∼R M ⊗R N.
=
With this, the proof becomes straightforward and is left to the reader.
Exercise: Complete the proof of Proposition 105.
Theorem 106. For a ﬂat algebra f : R → S , TFAE:
(i) S is faithfully ﬂat over R.
(ii) f ∗ : MaxSpec S → MaxSpec R is surjective.
(iii) f ∗ : Spec S → Spec R is surjective.
Proof. (i) ⇐⇒ (ii): Let m be any maximal ideal of R. Then mS
S holds iﬀ
there is a maximal ideal M of S containing mS iﬀ f ∗ (M) = m. The equivalence
now follows from criterion (iv) of Proposition 102.
(i) =⇒ (iii): Let p ∈ Spec R, and let k (p) be the fraction ﬁeld of the domain
R/p. By faithful ﬂatness, S ⊗R k (p) is a nonzero k (p)algebra so has a prime
f g ideal P . Consider the composite map h : R → S → S ⊗R k (p). We claim that
g ∗ : Spec(S ⊗R k (p)) → Spec R has image precisely {p}. The proof of this result, a
spectral description of the ﬁber of the morphism f : R → S over p, will have to COMMUTATIVE ALGEBRA 73 wait until we have developed the theory of localization in §7.3. Assuming it for now,
we get that g ∗ (P ) is a prime ideal of Spec S such that f ∗ g ∗ (P ) = (g ◦ f )∗ (P ) = p,
so f ∗ : Spec S → Spec R is surjective.
(iii) =⇒ (ii): Let m ∈ MaxSpec R ⊂ Spec R. By assumption, the set of prime
ideals P of S such that f ∗ P = m is nonempty. Moreover the union of any chain of
prime ideals pulling back to m is again a prime ideal pulling back to p, so by Zorn’s
Lemma there exists an ideal M which is maximal with respect to the property that
f ∗ M = m. Suppose M is not maximal and let M′ be a maximal ideal properly
containing M. Then by construction f ∗ (M′ ) properly contains the maximal ideal
m of R, i.e., f ∗ (M)′ ) = R, contradicting the fact that prime ideals pull back to
prime ideals. So M is indeed maximal in S .
Proposition 107. Let f : R → S be a ring extension such that S is a faithfully
ﬂat Rmodule, and let M be an Rmodule. Then:
a) M is ﬁnitely generated iﬀ M ⊗R S is ﬁnitely generated.
b) M is ﬁnitely presented iﬀ M ⊗R S is ﬁnitely presented.
Proof. Note ﬁrst that the properties of ﬁnite generation and ﬁnite presentation are
preserved by arbitrary base change f : R → S . So it suﬃces to prove that if M ⊗R S
is ﬁnitely generated (resp. ﬁnitely presented), then M is ﬁnitely generated (resp.
ﬁnitely presented).
a) Since M ⊗R S is ﬁnitely generated over S , it has a ﬁnite set of S module generators
of the form xi ⊗ 1. Let N = ⟨x1 , . . . , xn ⟩R and ι : N → M the canonical injection.
Then ιS : N ⊗R S → M ⊗R S is an isomorphism, so by faithful ﬂatness ι was itself
an isomorphism and thus M = ⟨x1 , . . . , xn ⟩ is ﬁnitely generated.
b) By part a), M is ﬁnitely generated over R, so let u : Rn → M be a surjection.
Since M ⊗R S is ﬁnitely presented, the kernel of uS : S n → M ⊗R S is ﬁnitely
generated over S . Since by ﬂatness ker uS = (ker u)S , part a) shows that ker u is
ﬁnitely generated and thus that M is ﬁnitely presented.
Lemma 108. Let f : R → S be a ring map, and let M, N be Rmodules.
a) There is a canonical S module map
ω : HomR (M, N ) ⊗R S → HomS (M ⊗R S, N ⊗R S )
such that for all u ∈ HomR (M, N ), ω (u ⊗ 1) = u ⊗ 1B .
b) If S is ﬂat over R and M is ﬁnitely generated, then ω is injective.
c) If S is ﬂat over R and M is ﬁnitely presented, then ω is an isomorphism.
Exercise: Prove Lemma ??. (It is not diﬃcult, really, but it is somewhat technical.
Feel free to consult [?, p. 23] for the details.)
Theorem 109. (Faithfully ﬂat descent for projective modules) Let f : R → S
be a faithfully ﬂat ring extension, and let P be an Rmodule. Then P is ﬁnitely
generated and projective iﬀ P ⊗R S is ﬁnitely generated and projective.
Proof. Begin, once again the implication P ﬁnitely generated projective implies
P ⊗R S is ﬁnitely generated projective holds for any base change. So suppose
P ⊗R S is ﬁnitely generated projective. Then P ⊗R S is ﬁnitely presented, so by
Proposition 107, M is ﬁnitely presented. It remains to show that M is projective.
Let v : M → N be a surjection of Rmodules. We wish to show that the natural
map HomR (P, M ) → HomR (P, N ) is surjective. Because of the faithful ﬂatness 74 PETE L. CLARK of S/R, it is suﬃcient to show that HomR (P, M ) ⊗R S → HomR (P, N ) ⊗R S is
surjective, and by Lemma 108 this holds iﬀ
HomS (P ⊗R S, M ⊗R S ) → HomS (P ⊗R S, N ⊗R S )
is surjective. But this latter map is surjective because M ⊗R S → N ⊗R S is
surjective and the S module P ⊗R S is projective by assumption.
4. First Properties of Ideals in a Commutative Ring
4.1. Introducing maximal and prime ideals.
Consider again the set I (R) of all ideals of R, partially ordered by inclusion. The
maximal element is the ideal R itself, and the minimal element is the ideal (0).
In general our attitude to the ideal R of R is as follows: although we must grudgingly admit its existence – otherwise, given a subset S of R it would be in general
a diﬃcult question to tell whether the ideal ⟨S ⟩ generated by S “exists” (i.e., is
proper) or not – nevertheless we regard it as exceptional and try to ignore it as
much as possible. Because of this we deﬁne an ideal I of R to be maximal if it is
maximal among all proper ideals of R, i.e., I R and there does not exist J such
that I J R. That this is a more interesting concept than the literally maximal
ideal R of R is indicated by the following result.
Proposition 110. For an ideal I of R, TFAE:
(i) I is maximal.
(ii) R/I is a ﬁeld.
Proof. Indeed, R/I is a ﬁeld iﬀ it has precisely two ideals, I and R, which by
the Correspondence Theorem says precisely that there is no proper ideal strictly
containing I .
Example: In R = Z, the maximal ideals are those of the form (p) for p a prime
number. The quotient Z/pZ is the ﬁnite ﬁeld of order p.
Does every ring have a maximal ideal? With a single (trivial) exception, the answer
is yes, assuming – as we must, in order to develop the theory as it is used in other
branches of mathematics – suitable transﬁnite tools.
Proposition 111. Let R be a nonzero ring and I a proper ideal of R. Then there
exists a maximal ideal of R containing I .
Proof. Consider the set S of all proper ideals of R containing I , partially ordered
by inclusion. Since I ∈ S , S is nonempty. Moreover the union of a chain of ideals
is an ideal, and the union of a chain of proper ideals is proper (for if 1 were in the
union, it would have to lie in one of the ideals of the chain). Therefore by Zorn’s
Lemma we are entitled to a maximal element of S , which is indeed a maximal ideal
of R that contains I .
Corollary 112. A nonzero ring R contains at least one maximal ideal.
Proof. Apply Proposition 111 with I = (0). COMMUTATIVE ALGEBRA 75 Remark: The zero ring has the disquieting property of having no maximal ideals.
Remark: The appeal to Zorn’s Lemma cannot be avoided, in the sense that Corollary 112 implies the Axiom of Choice.
A proper ideal I of a ring R is prime if xy ∈ I implies x ∈ I or y ∈ I .
Exercise 4.1: Let p be a prime ideal of R.
a) Suppose x1 , . . . , xn are elements of R such that x1 · · · xn ∈ p. Then xi ∈ p for
some at least one i.
b) In particular, for x ∈ R and n ∈ Z+ we have xn ∈ p, then x ∈ p.
Proposition 113. Let f : R → S be a homomorphism of rings, and let J be an
ideal of S .
a) Put f ∗ (J ) := f −1 (J ) = {x ∈ R  f (x) ∈ J }. Then f ∗ (J ) is an ideal of R.
b) If J is a prime ideal, so is f ∗ (J ).
Exercise 4.2: Prove Proposition 113.
Proposition 114. For a commutative ring R, TFAE:
(i) If x, y ∈ R are such that xy = 0, then x = 0 or y = 0.
(ii) If 0 ̸= x ∈ R and y, z ∈ R are such that xy = xz , then y = z .
A ring satisfying either of these two properties is called an integral domain.
Proof. Assume (i), and consider xy = xz with x ̸= 0. We have x(y − z ) = 0, and
since x ̸= 0, (i) implies y − z = 0, i.e., y = z . Assuming (ii) suppose xy = 0 with
x ̸= 0. Then xy = 0 = x · 0, so applying cancellation we get y = 0.
A zero divisor in a ring R is an element x such that there exists 0 ̸= y ∈ R with
xy = 0. (In particular 0 is a zero divisor, albeit not a very interesting one.) So
property (i) expresses that there are no zero divisors other than 0 itself. Property
(ii) makes sense in any commutative monoid and is called cancellation.
Remark: The terminology “integral domain” is motivated by the fact the integers
Z satisfy (i) and (ii) of Proposition 114, so any ring which satisﬁes these properties
can be viewed as a sort of ring of “generalized integers.” The analogy is apt – in
particular, later on we shall build from any integral domain a ﬁeld of fractions in
exactly the same way that the rational numbers are constructed from the integers
– but the terminology is quite awkward, as the reader will come to appreciate.
First of all the word “integral” is logically superﬂuous – we do not have any other
deﬁnition of a domain, and indeed often we will use domain as a more succinct
synonym for “integral domain”. So why not just “domain”? One problem is that,
being a property of an object and not an object itself, we would prefer an English
word which is an adjective rather than a noun. So why not just “integral”? The
problem is that the word “integral” will be used later for something else (not a
property of a single ring but a property of an extension ring S of R). It would certainly be confusing to use the term integral for these two diﬀerent concepts. Ideal
would probably be to reserve the term “integral” for a ring without zero divisors,
give some other name to integral extensions (integrally algebraic?), and eliminate the use of “domain” from terms like “principal ideal domain.” (This would be
consistent with geometric terminology: an aﬃne scheme Spec R is called integral 76 PETE L. CLARK iﬀ R is an integral domain.) However, the terminology is too entrenched for this to
be a feasible solution.
Proposition 115. For an ideal I in a ring R, TFAE:
(i) I is prime.
(ii) R/I is an integral domain.
Exercise 4.3: Prove Proposition 115.
Corollary 116. A maximal ideal is prime.
Proof. If I is maximal, R/I is a ﬁeld, hence an integral domain, so I is prime.
Corollary 116 is the ﬁrst instance of a somewhat mysterious metaprinciple in ideal
theory: for some property P of ideals in a ring, it is very often the case that an
ideal which is maximal with respect to the satisfaction of property P (i.e., is not
strictly contained in any other ideal satisfying P) must be prime. In the above, we
saw this with P = “proper”. Here is another instance:
Proposition 117. (Multiplicative Avoidance) Let R be a ring and S ⊂ R. Suppose: 1 is in S ; 0 is not in S ; and S is closed under multiplication: S · S ⊂ S .
Let IS be the set of ideals of R which are disjoint from S . Then:
a) IS is nonempty;.
b) Every element of IS is contained in a maximal element of IS .
c) Every maximal element of IS is prime.
Proof. a) (0) ∈ IS . b) Let I ∈ IS . Consider the subposet PI of IS consisting of
ideals which contain I . Since I ∈ PI , PI is nonempty; moreover, any chain in PI
has an upper bound, namely the union of all of its elements. Therefore by Zorn’s
Lemma, PI has a maximal element, which is clearly also a maximal element of IS .
c) Let I be a maximal element of IS ; suppose that x, y ∈ R are such that xy ∈ I .
If x is not in I , then ⟨I, x⟩ I and therefore contains an element s1 of S , say
s1 = i1 + ax.
Similarly, if y is not in I , then we get an element s2 of S of the form
s2 = i2 + by.
But then
s1 s2 = i1 i2 + (by )i1 + (ax)i2 + (ab)xy ∈ I ∩ S,
a contradiction.
In fact Corollary 116 is precisely the special case S = {1} of Proposition 117.
If I and J are ideals of R, we deﬁne the product IJ to be the ideal generated
by all∑
elements of the form xy with x ∈ I, y ∈ J . Every element of IJ is of the
n
form i=1 xi yi with x1 , . . . , xn ∈ I, y1 , . . . , yn ∈ J .
The following simple result will be used many times in the sequel.
Proposition 118. Let p be a prime ideal and I1 , . . . , In be ideals of a ring R. If
p ⊃ I1 · · · In , then p ⊃ Ii for at least one i.
Proof. An easy induction argument reduces us to the case of n = 2. So suppose
for a contradiction that p ⊃ I1 I2 but there exists x ∈ I1 \ p and y ∈ I2 \ p. Then
xy ∈ I1 I2 ⊂ p; since p is prime we must have x ∈ p or y ∈ p, contradiction. COMMUTATIVE ALGEBRA 77 Exercise 4.4: Show that Proposition 118 characterizes prime ideals, in the sense
that if p is any ideal such that for all ideals I, J of R, p ⊂ IJ implies p ⊂ I or
p ⊂ JJ , then p is a prime ideal.
For an ideal I and n ∈ Z+ , we denote the nfold product of I with itself by I n .
Corollary 119. If p is a prime ideal and I is any ideal, then p ⊃ I n =⇒ p ⊃ I .
4.2. Radicals.
An element x of a ring R is nilpotent if xn = 0 for some n ∈ Z+ . Obviously
0 is a nilpotent element; a ring in which 0 is the only nilpotent element is called
reduced. An ideal I of R is nil if every element of I is nilpotent. An ideal I is
nilpotent if there exists n ∈ Z+ such that I n = (0).
Proposition 120. Let I be an ideal of a ring R.
a) If I is nilpotent, then I is a nil ideal.
b) If I is ﬁnitely generated and nil, then I is nilpotent.
Proof. Part a) is immediate from the deﬁnition, as we invite the reader to check.
Suppose I = ⟨a1 , . . . , an ⟩R . Since I is nil, for each i, 1 ≤ i ≤ r, there exists ni
such that ani = 0. Let N = n1 + . . . + nr . We claim I N = 0. Indeed, an arbirary
i
element of I is of the form x1 a1 + . . . + xn an . Raising this element to the nth power
∑r
yields a sum of monomials of the form xj1 · · · xjr aj1 · · · ajr , where i=1 ji = N . If
r
r1
1
we had for all i that ji < ni , then certainly j1 + . . . + jr < N . So for at least one i
we have ji ≥ ni and thus xji = 0; so every monomial term equals zero.
i
Exercise 4.5: Find a ring R and an ideal I of R which is nil but not nilpotent.
The nilradical N of R is the set of all nilpotent elements of R.
Proposition 121. Let R be a ring.
a) The nilradical N is a nil ideal of R.
b) The quotient R/N is reduced.
c) The map q : R → R/N is universal for maps from R into a reduced ring.
d) The nilradical is the intersection of all prime ideals of R.
Proof. a) In establishing that N is an ideal, the only property which is not absolutely immediate is its closure under addition. Suppose xm = 0 = y n . Then every
term in the binomial expansion of (x + y )m+n−1 is an integer times xi y m+n−1−i
for 0 ≤ i ≤ m + n. Here either i ≥ m so xi y m+n−1−i = 0 · y m+n−i = 0, or
m + n − 1 − i ≥ n, so xi y m+n−i−i = xi · 0 = 0, so x + y is nilpotent and N is an
ideal, and by deﬁnintion a nil ideal.
b) Let r + N be a nilpotent element of R/N , so there exists n ∈ Z+ such that
rn ∈ N . But this means there exists m ∈ Z+ such that 0 = (rn )m = rnm , and thus
r itself is a nilpotent element.
c) In plainer terms: if S is a reduced ring and f : R → S is a ring homomorphism,
then there exists a unique homomorphism f : R/N → S such that f = f ◦ q . Given
this, the proof is straightforward, and we leave it to the reader.
d) Suppose x is a nilpotent element of R, i.e., ∃n ∈ Z+ such that xn = 0. If ∩ is
p
a prime ideal, then since 0 = x · · · x ∈ p, we conclude x ∈ p: this shows N ⊂ p.
Conversely, suppose x is not nilpotent. Then the set Sx := {xn  n ∈ N} satisﬁes 78 PETE L. CLARK (i) and (ii) of Proposition 117, so we may apply that result to get a prime ideal p
which is disjoint from Sx , hence not containing x.
Exercise 4.6: Prove Proposition 121c).
An ideal I of a ring R is radical if for all x ∈ R, n ∈ Z+ , xn ∈ I implies x ∈ I .
Exercise 4.7: a) Show that a prime ideal is radical.
b) Exhibit a radical ideal which is not prime.
c) Find all radical ideals in R = Z.
d) Show that R is reduced iﬀ (0) is a radical ideal.
∩
e) Let {Ii } be a set of radical ideals in a ring R. Show I = i Ii is a radical ideal.
For any ideal I of R, we deﬁne the radical of I :
r(I ) = {x ∈ R  ∃n ∈ Z+ xn ∈ I }.
Proposition 122. Let R be a commutative ring and I , J ideals of R.
a) r(I ) is the intersection of all prime ideals containing I , and is a radical ideal.
b) (i) I ⊂ r(I ), (ii) r(r(I )) = r(I ); (iii) I ⊂ J =⇒ r(I ) ⊂ r(J ).
c) r(IJ ) = r(I ∩ J ) = r(I ) ∩ r(J ).
d) r(I + J ) = r(r(I ) + r(J )).
e) r(I ) = R ⇐⇒ I = R.
f ) For all n ∈ Z+ , r(I n ) = r(I ).
Proof. First we make the following observation: under the canonical homomorphism q : R → R/I , r(I ) = q −1 (N (R/I )). By Proposition 113a), r(I ) is an ideal.
a) Since N is the intersection of all prime ideals of R/I , r(I ) is the intersection of
all prime ideals containing I , which is, by Exericse X.Xe), a radical ideal.
b) (i) is immediate from the deﬁnition, and (ii) and (iii) follow from the characterization of r(I ) as the intersection of all radical ideals containing I .
c) Since IJ ⊂ I ∩ J , r(IJ ) ⊂ r(I ∩ J ). If xn ∈ I ∩ J , then x2n = xn xn ∈ IJ , so
x ∈ r(IJ ); therefore r(IJ ) = r(I ∩ J ). Since I ∩ J is a subset of both I and J ,
r(I ∩ J ) ⊂ r(I ) ∩ r(J ). Conversely, if x ∈ r(I ) ∩ r(J ), then there exist m and n
such that xm ∈ I and xn ∈ J , so xmn ∈ I ∩ J and x ∈ r(I ∩ J ).
d) Since I + J ⊂ r(I ) + r(J ), r(I + J ) ⊂ r(r(I ) + r(J )). A general element of
r(I )+ r(J ) is of the form x + y , where xm ∈ I and y n ∈ J . Then (x + y )m+n ∈ I + J ,
so x + y ∈ r(I + J ).
e) Evidently r(R) = R. Conversely, if r(I ) = R, then there exists n ∈ Z+ such that
1 = 1n ∈ I .
f) By part a), r(I n ) is the intersection of all prime ideals p ⊃ I n . But by Corollary
119, a prime contains I n iﬀ it contains I , so r(I n ) = r(I ).
Remark: Proposition 122b) asserts that the mapping I → r(I ) is a closure operator on the lattice I (R) of ideals of R.
Exercise 4.8: Let I be an ideal in the ring R. Show that r(I ) is the intersection of all prime ideals containing I . (Hint: reduce to the case I = 0.)
An ideal p of a ring R is primary if every zero divisor of R/p is nilpotent. Equivalently, xy ∈ p, x ∈ p =⇒ y n ∈ p for some n ∈ Z+ . More on primary ideals in §X.X .
/ COMMUTATIVE ALGEBRA 79 We also deﬁne the Jacobson radical J (R) as the intersection of all maximal
ideals of R. Evidently we have N ⊂ J (R).
Proposition 123. Let R be a ring. An element x of R lies in the Jacobson radical
J (R) iﬀ 1 − xy ∈ R× for all y ∈ R.
Proof. Suppose x lies in every maximal ideal of R. If there exists y such that 1 − xy
is not a unit of R, then 1 − xy lies in some maximal ideal m, and then x ∈ m implies
xy ∈ m and then 1 = (1 − xy ) + xy ∈ m, a contradiction. Conversely, suppose that
there is a maximal ideal m of R which does not contain x. Then ⟨m, x⟩ = R, so
1 = m + xy for some m ∈ m and y ∈ R, and thus 1 − xy is not a unit.
Remark: It is not yet clear why we have deﬁned these two diﬀerent notions of
“radical.” Neither is it so easy to explain in advance, but nevertheless let us make
a few remarks. First, the Jacobson radical plays a very important role in the theory of noncommutative rings, especially that of ﬁnite dimensional algebras over a
ﬁeld. (Indeed, a ﬁnite dimensional k algebra is semisimple – i.e., a direct product
of algebras without nontrivial twosided ideals – iﬀ its Jacobson radical is zero.
In the special case of commutative algebras this comes down to the simpler result
that a ﬁnite dimensional commutative k algebra is reduced iﬀ it is a product of
ﬁelds.) Note that one important place in commutative algebra in which the Jacobson radical J (R) appears – albeit not by name, because of the necessity of putting
the results in a ﬁxed linear order – is in the statement of Nakayama’s Lemma. In
general, the deﬁning condition of nil(R) – i.e., as the intersection of all prime ideals
of R – together with the fact that the radical of an arbitrary ideal I corresponds
to the nilradical of R/I , makes the nilradical more widely useful in commutative
algebra (or so it seems to the author of these notes). It is also important to consider
when the nil and Jacobson radicals of a ring coincide. A ring R for which every
homomorphic image S has nil(S ) = J (S ) is called a Jacobson ring; such rings
will be studied in detail in §X.X .
4.3. Comaximal ideals.
Recall (from Section X.X) that two ideals I and J in a commutative ring R are
comaximal if I + J = R.
Exercise 4.9: Determine all pairs of comaximal ideals in R = Z.
Proposition 124. Let I and J be ideals in R. If r(I ) and r(J ) are comaximal, so
are I and J .
Proof. Apply Proposition 122d) and then Proposition 122e) to r(I ) + r(J ) = R:
R = r(r(I ) + r(J )) = r(I + J ) = I + J.
Recall that a set {Ii } of ideals of R is pairwise comaximal if for each i ̸= j ,
Ii + Ij = R. An immediate corollary of Proposition 124 is that if {Ii } are pairwise
n
comaximal and {ni } are any positive integers, then {Ii i } are pairwise comaximal.
Lemma 125. Let J ,K1 , . . . , Kn be pairwise comaximal ideals in the ring R. Then
∩n
K1 · · · Kn = i=1 Ki and J + K1 · · · Kn = R. 80 PETE L. CLARK Proof. We go by induction on n, the case of n = 1 being Lemma 26b). Suppose
the theorem is true for any collection of n − 1 ideals K1 , . . . , Kn−1 and an ideal
M such that the set {K1 , . . . , Kn−1 , M } is pairwise comaximal. First apply the
∩n−1
induction hypothesis with M = Kn to get that K ′ := K1 · · · Kn−1 = i=1 Ki and
K ′ + Kn = R. Now part b) gives that
K1 · · · Kn = K = K ′ Kn = K ′ ∩ Kn = n
∩ Ki . i=1 Now apply the induction hypothesis with M = J to get that K ′ = K1 · Kn−1 and
J are comaximal, so so that there exist j ∈ J , k ′ ∈ K ′ such that j + k ′ = 1. Now
let kn be any element of kn ; we have kn = jkn + k ′ kn ∈ J + K . Therefore J + K
contains both J and Kn , but since J + Kn = R this implies J + K = R.
Theorem 126. (Chinese Remainder Theorem, or “CRT”) Let R be a ring and
I1 , . . . , In a ﬁnite set of pairwise comaximal ideals. Consider the natural map
n
∏
Φ:R→
R/Ii ,
i=1 x → (x + Ii )n . Then Φ is surjective with kernel I1 · · · In , so that there is an
i=1
induced isomorphism
n
∏
∼
Φ : R/(I1 · · · In ) →
R/Ii .
i=1 ∩n
Proof. The map Φ is welldeﬁned and has kernel i=1 Ii . Since the Ii ’s are pairwise
∩n
comaximal, Lemma 125 gives i=1 Ii = I1 · · · In . So it remains to show that Φ is
surjective. We prove this by induction on n, the case n = 1 being trivial. So we
∏n−1
may assume that the natural map Φ′ : R → R′ := i=1 R/Ii is surjective, with
kernel I ′ := I1 · · · In−1 . Let (r′ , s) be any element of R′ × R/In . By assumption,
there exists r ∈ R such that Φ′ (r + I ′ ) = r′ . Let s be any element of R mapping
to s ∈ R/In . By Lemma 125, I ′ + In = R, so there exist x ∈ I ′ , y ∈ In such that
s − r = x + y . Then Φ′ (r + x) = r′ , and r + x ≡ r + x + y ≡ s (mod In ), so
Φ(r + x) = (r′ , s) and Φ is surjective.
Remark: In the classical case R = Z, we can write Ii = (ni ) and then we are
trying to prove – under the assumption that the ni ’s are coprime in pairs in
the sense of elementary number theory – that the injective ring homomorphism
Z/(n1 · · · nn ) → Z/n1 × · · · × Z/nn is an isomorphism. But both sides are ﬁnite
rings of order n1 · · · nn , so since the map is an injection it must be an isomorphism!
Nevertheless the usual proof of CRT in elementary number theory is much closer
to the one we gave in the general case: in particular, it is constructive.
Exercise 4.10: Let∏ be a ring and I1 , . . . , In any ﬁnite sequence of ideals. Consider
R
n
the map Φ : R → i=1 R/Ii as in CRT.
a) Show that Φ is surjective only if the {Ii } are pairwise comaximal.
∩n
b) Show that Φ is injective iﬀ i=1 Ii = (0).
Exercise 4.11:
a) Let G be a ﬁnite commutative group with exactly one element z of order 2. Show
∑
that x∈G x = z . COMMUTATIVE ALGEBRA 81 b) Let G be a ﬁnite commutative group which does not have exactly one element
∑
of order 2. Show that x∈G x = 0.
c) Prove the following result of Gauss (a generalization of Wilson’s Theorem):
let N ∈ Z+ , and put
∏
P (N ) =
x.
x∈(Z/N Z)× Then: P (N ) = ±1, and the minus sign holds iﬀ N = 4 or is of the form pm or 2pm
for an odd prime p and m ∈ Z+ .
d) For a generalization to the case of (ZK /A)× , where A is an ideal in the ring ZK
of integers of a number ﬁeld K , see [Da09]. Can you extend Dalawat’s results to
the function ﬁeld case?
Exercise 4.12: Let K be any ﬁeld, and put R = K [t].
a) Let n1 , . . . , nk be a sequence of nonnegative integers and {x1 , . . . , xk } a k element subset of K . For 1 ≤ i ≤ k , let ci0 , . . . , cini be a ﬁnite sequence of ni + 1
elements of k (not necessarily distinct). By applying the Chinese Remainder Theorem, show that there is a polynomial P (t) such that for 1 ≤ i ≤ k and 0 ≤ j ≤ ni
we have P (j ) (xi ) = cij , where P (j ) (xi ) denotes the j th “formal” derivative of P
evaluated at xi . Indeed, ﬁnd all such polynomials; what can be said about the least
degree of such a polynomial?
b) Use the proof of the Chinese Remainder Theorem to give an explicit formula for
such a polynomial P .
Exercise 4.13: Let (M, ·) be a monoid and k a ﬁeld. A character on M with
values in k is a homomorphism of monoids from M to the multiplicative group k ×
of k . Each character lies in the k vector space k M of all functions from M to k .
a) (Dedekind) Show that any ﬁnite set of characters is k linearly independent.
b) Give an example of an inﬁnite set of characters which is k linearly dependent.
Exercise 4.14: Show that for a ring R, TFAE:
(i) R has ﬁnitely many maximal ideals.19
(ii) The quotient of R by its Jacobson radical J (R) is a ﬁnite product of ﬁelds.
Exercise 4.15: Let R be a ring with the property that for every nonzero ideal I
of R, R/I is ﬁnite.20
a) Show that every nonzero prime ideal of R is maximal.
b) For a nonzero ideal I of R, deﬁne N (I ) = #R/I . Note that CRT implies that
N is multiplicative in the sense that if I and J are comaximal then N (IJ ) =
N (I )N (J ). In particular, show that this multiplicativity holds for any distinct
nonzero prime ideals p and q.
c) Show that for any n ∈ Z+ there are only ﬁnitely many ideals I with N (I ) ≤ n.
We can therefore deﬁne a formal Dirichlet series
∑
ζR (s) =
N (I )−s ,
I { 0}
19Such rings are typically called semilocal. I am not a fan of the terminology – it seems to
either suggest that R has one half a maximal ideal (whatever that could mean) or two maximal
ideals. But it is well entrenched, and I will not campaign to change it.
20We say that R has ﬁnite quotients. 82 PETE L. CLARK the Dedekind zeta function of R.
d) Suppose that every nonzero ideal I of R factors uniquely into a product of
prime ideals.21 Show that N is completely multiplicative in the sense that for
all nonzero ideals I and J of R, N (IJ ) = N (I )N (J ). Deduce an Euler product
factorization for ζR (s).
4.4. Local rings. All rings are commutative with unity.
Proposition 127. For a ring R, TFAE:
(i) There is exactly one maximal ideal m.
(ii) The set R \ R× of nonunits forms a subgroup of (R, +).
(iii) The set R \ R× is a maximal ideal.
A ring satisfying these equivalent conditions is called a local ring.
∪
Proof. Since R× = R \ m m, the union extending over all maximal ideals of R, it
follows that if there is only one maximal ideal m then m = R \ R× . This shows
(i) =⇒ (iii) and certainly (iii) =⇒ (ii). Conversely, since the set of nonunits of
a ring is a union of ideals, it is closed under multiplication by all elements of the
ring. Thus it is itself an ideal iﬀ it is an additive subgroup: (ii) =⇒ (iii). The
implication (iii) implies (i) is very similar and left to the reader.
Warning: In many older texts, a ring with a unique maximal ideal is called “quasilocal” and a local ring is a Noetherian quasilocal ring. This is not our convention.
Local rings (especially Noetherian local rings) play a vital role in commutative algebra: the property of having a single maximal ideal simpliﬁes many idealtheoretic
considerations, and many ring theoretic considerations can be reduced to the study
of local rings (via a process called, logically enough, localization : see §X ).
A ﬁeld is certainly a local ring. The following simple result builds on this trivial observation to give some further examples of local rings:
Proposition 128. Let I be an ideal in the ring R.
a) If rad(I ) is maximal, then R/I is a local ring.
b) In particular, if m is a maximal ideal and n ∈ Z+ then R/mn is a local ring.
∩
Proof. a) We know that rad(I ) = p⊃I p, so if rad(I ) = m is maximal it must be
the only prime ideal containing I . Therefore, by correspondence R/I is a local ring.
(In fact it is a ring with a unique prime ideal.)
b) By Proposition 122f), r(mn ) = r(m) = m, so part a) applies.
So, for instance, for any prime number p, Z/(pk ) is a local ring, whose maximal
ideal is generated by p. It is easy to see (using, e.g. the Chinese Remainder Theorem) that conversely, if Z/(n) is a local ring then n is a prime power.
Example: The ring Zp of padic integers is a local ring. For any ﬁeld k , the ring
k [[t]] of formal power series with coeﬃcients in k is a local ring. Both of these rings
are also PIDs. A ring which is a local PID is called a discrete valuation ring;
these especially simple and important rings will be studied in detail later.
21We will see later that this is a class of rings called Dedekind rings which can be characterized in many other ways. COMMUTATIVE ALGEBRA 83 Exercise 4.16: Show that a local ring is connected, i.e., e2 = e =⇒ e ∈ {0, 1}.
5. Examples of rings
5.1. Rings of numbers. The most familiar examples of rings are probably rings
of numbers, e.g.
Z ⊂ Q ⊂ R ⊂ C.
These are, respectively, the integers, the rational numbers, the real numbers and
the complex numbers. For any positive integer N the ring integers modulo N , denoted Z/N Z. We assume that the reader has seen all these rings before.
Historically, the concept of a ring as an abstract structure seems to have arisen
as an attempt formalize common algebraic properties of number rings of various
sorts. It is my understanding that the term “ring” comes from Hilbert’s Zahlring
(“Zahl” means “number” in German). Indeed, various sorts of extension rings of
C – most famously Hamilton’s quaternions H – have been referred to as systems
of hypercomplex numbers. This terminology seems no longer to be widely used.
The adjunction process gives rise to many rings and ﬁelds of numbers, as already
√
seen in §2.2. For instance, for any nonsquare integer D, let D be a complex
√
number whose square is D: then Z[ D] is an interesting ring.
√
√
Exercise 5.1: Show that Z[ D] = {a + b D  a, b ∈ Z}.
√
In√
particular, (Z[ D], +) ∼ (Z2 , +) as abelian groups, although not as rings, since
=
Z[ D] is an integral domain and Z2 , being a disconnected ring, has nontrivial
idempotents.
More generally, let K be any number ﬁeld, i.e., a ﬁnite degree ﬁeld extension of Q.
Deﬁne ZK to be the set of elements x ∈ K which satisfy a monic polynomial with
Zcoeﬃcients. It turns out that ZK is a ring, the ring of algebraic integers in
K. This is a very special case of the theory of integral closure, which we will study
in great detail later on.
Algebraic number theory proper begins with the observation that in general the
rings ZK need not be UFDs but are otherwise as nice as possible from a commutative algebraic standpoint. That is, every ring ZK is a Dedekind domain, which
among many other characterizations, means that every nonzero ideal factors into a
product of prime ideals. That the rings ZK are Dedekind domains is an example of
a normalization theorem, more speciﬁcally a very special case of the KrullAkizuki
Theorem, which we will meet later on.
In particular, if one wanted to give an extremely laconic description of algebraic
number theory – certainly incomplete, but rather accurate in spirit – one could say
that it is the study of the Picard groups Pic ZK of rings of integers of number ﬁelds.
Let Q be an algebraic closure of Q. (This is not a number ﬁeld, being an inﬁnite degree algebraic extension of Q.) We may deﬁne Z to be the set of all elements 84 PETE L. CLARK of Q which satsify a monic polynomial with integer coeﬃcients: this is the ring of
all algebraic integers. In particular,
Z = lim ZK
−
→
is the direct limit of all rings of integers in ﬁxed number ﬁelds. This is a nice example of a ring which is natural enough but is “bigger” than the rings which were
more classically studied.
Exercise 5.2: Let Z be the set of all algebraic integers.
a) Taking as given that for any ﬁxed number ﬁeld K , the algebraic integers in K
form a subring of K , show that Z is a subring of Q.
b) Show that Z is an integral domain which is not Noetherian. Hint: use the fact
that the nth root of an algebraic integer is an algebraic integer to construct an
inﬁnite strictly ascending chain of principal ideals in Z.
Theorem 129. Every ﬁnitely generated ideal in the ring Z is principal.
Thus, if only Z were Noetherian, it would be a principal ideal domain! Later on
we will prove a more general theorem, due to Kaplansky, in the context of limits of
Dedekind domains with torsion Picard groups.
5.2. Rings of continuous functions.
Let R be a ring, X a set, and consider the set RX of all functions f : X → R. We
may endow RX with the structure of a ring by deﬁning addition and multiplication
“pointwise”, i.e.,
(f + g ) : x → f (x) + g (x),
(f g ) : x → f (x)g (x).
Exercise X.X: Show that this makes RX into a ring with additive identity the constant function 0 and multiplicative identity the constant function 1.
However, this is not really a “new” example of a ring.
Exercise 5.3: Show that RX is isomorphic as a ring, to ∏
x ∈X R. Later on we will see that this sort of construction is of some use in the special
case that R = F , in which case we get an important class of Boolean rings. However, in general RX is quite a roomy ring. It contains many interesting subrings,
some of which can be nicely consructed and analyzed using topological, geometric
and analytic considerations.
Suppose instead that we specialize to the following situation: R = R (the real
numbers!), X is not just any old set but a topological space, and instead of the ring
RX of all functions f : X → R we look instead at the subring C (X ) of continuous
functions.
Exercise 5.4: Show that for a topological space X , TFAE:
(i) For every x, y ∈ X with x ̸= y , there exists f ∈ C (X ) with f (x) ̸= f (y ).
(ii) For every x, y ∈ X with x ̸= y and every α, β ∈ R, there exists f ∈ C (X ) with
f (x) = α, f (y ) = β . COMMUTATIVE ALGEBRA 85 (iii) For every ﬁnite subset S of X and any function g : S → R, there exists
f ∈ C (X ) such that f S = g .
A space which satisﬁes these equivalent conditions is called Cseparated.22
Recall the following chains of implications from general topology:
Lemma 130. For any topological space, the following implications hold (and none
of the arrows may be reversed)):
a) X compact =⇒ X normal =⇒ X Tychonoﬀ =⇒ X regular =⇒ X
Hausdorﬀ =⇒ X separated =⇒ X Kolmogorov.
b) X locally compact =⇒ X Tychonoﬀ.
Exercise 5.5: a) Show that a Tychonoﬀ space is Cseparated.
b) Show that a Cseparated space is Hausdorﬀ.
c)* Show that a regular space need not be Cseparated. (Suggestion: see [Ga71].)
For a topological space X , a zero set is a set of the form f −1 (0) for some continuous function f : X → R. A cozero set is a complement of a zero set. The
cozero sets in fact form a base for a topology on X , called (by us, at least) the
Ztopology. Let us write XZ for X endowed with the Z topology. Since every
cozero set is an open set in the given topology on X , XZ is a coarser topology than
the given topology on X : of course by this we allow the possibility that the two
topologies coincide: i.e., every closed set is an intersection of zero sets of continuous
Rvalued functions. The following basic (but not so widely known) result gives a
condition for this.
Theorem 131. a) For a Hausdorﬀ topological space X , the following are equivalent:
(i) XZ = X : every closed set is an intersection of zero sets of continuous functions.
(ii) X is Tychonoﬀ, i.e., if Y is a closed subset of X and x ∈ X \ Y , then there
exists a continuous function f : X → [0, 1] with f (x) = 0, f Y ≡ 1.
b) For any topological space X , the space XZ is completely regular, and is the ﬁnest
completely regular topology on the underlying set of X which is coarser than X .
Proof. GilmanJerison, p. 38.
Let X be a topological space, and let x ∈ X be any point. Consider the set
mx = {f ∈ C (X )  f (x) = 0}.
Evidently mx is an ideal of C (X ). But more is true.
∼ Proposition 132. Evaluation at x gives a canonical isomorphism C (X )/mx → R.
In particular, mx is a maximal ideal of C (X ).
Exercise 5.6: Prove Proposition 132.
Thus x → mx gives a map of sets M : X → M (X ).
Proposition 133. The map M : X → M (X ) is injective iﬀ X is Cseparated.
Proof. This is left to the reader as a routine check on parsing the deﬁnitions.
22More standard terminology: “the continuous functions on X separate points”. 86 PETE L. CLARK Proposition 134. If X is quasicompact, then M is surjective, i.e., every maximal
ideal of C (X ) is of the form mx for at least one point x ∈ X .
Proof. It suﬃces to show: let I be an ideal of C (X ) such that for no x ∈ X do we
have I ⊂ mx . Then I = C (X ).
By hypotehsis, for every x ∈ X there exists fx ∈ I such that fx (x) ̸= 0. Since
fx is continuous, there exists an open neighborhood Ux of x such fx is nowhere
vanishing on Ux . By quasicompactness of X , there exists a ﬁnite set x1 , . . . , xN
∪N
2
2
such X = i=1 Uxi . Then the function f = fx1 + . . . + fxn is an element of m which
1
is strictly positive at every x ∈ X . But then f is also a continuous function on X ,
i.e., f ∈ C (X )× , so I = R.
A compact space is quasicompact and Cseparated. Thus previous results yield:
Theorem 135. If X is a compact space, then M : X → M (X ) is a bijection:
every maximal ideal of the ring of continuous functions on X is of the form mx for
a unique x ∈ X .
In fact more is true. The perspective of functional analysis yields a natural topology
on M (X ), the initial topology: namely, each f ∈ C (X ) induces a function Mf :
M (X ) → R, namely Mf maps m to the image of f in C (X )/m = R. Now we
endow M (X ) with the coarsest topology which makes each of the functions Mf
continuous.
Lemma 136. For a compact space X , the initial topology on M (X ) is Hausdorﬀ.
Proof. For distinct x, x′ ∈ X , consider the maximal ideals mx , mx′ . By Cseparatedness,
there exists f ∈ C (X ) with f (x) = 0, f (x′ ) ̸= 0. Thus choose disjoint neighborhoods V, V ′ of f (x), f (x′ ) ∈ R. The sets
Uf,V = {x ∈ X  f (x) ∈ V }, Uf,V ′ = {x ∈ X  f (x) ∈ V ′ }
are disjoint open neighborhoods of x and x′ .
Theorem 137. For a compact space X , let M (X ) be the set of maximal ideals
of C (X ) endowed with the initial topology. Then M : X → M (X ), x → mx is a
homeomorphism.
Proof. Step 1: We claim that M is continuous. But indeed, by the universal
property of the initial topology, it is continuous iﬀ for all f ∈ C (X ), the composite
function x → mx → f (mx ) is continuous. But this is nothing else than the function
x → f (x), i.e., the continuous function f ! So that was easy.
Step 2: We now know that M is a continuous bijection from a compact space to
a Hausdorﬀ space. Therefore it is a closed map: if Y is closed in X , then Y is
compact, so M(Y ) is compact, so M(Y ) is closed. Therefore M−1 is continuous
and thus M is a homeomorphism.
For any commutative ring R, we deﬁne MaxSpec(R) to be the maximal ideals and
put a topology on it: for any ideal I of R, we deﬁne
V (I ) = {m ∈ MaxSpec R  I ⊂ m}
The sets V (I ) are the closed sets for a unique topology on MaxSpec R, the Zariski
topology. Another way to say it is that the closed sets in the Zariski topology are
precisely all sets obtained by intersecting sets of the form
V (f ) = {m ∈ MaxSpec R  f ∈ m}. COMMUTATIVE ALGEBRA 87 To see this, note ﬁrst that for any ideal I of R,
∩
V (I ) =
V (f )
f ∈I and for any subset S of R, ∩
f ∈S V (f ) = ∩ V (f ). f ∈⟨S ⟩R Thus, from the perspective of the rest of these notes, it is natural to consider
M (X ) = MaxSpec C (X ) as being endowed with the Zariski topology rather than
the initial topology (note that the latter is deﬁned only in the quasicompact case).
Proposition 138. Let X be any topological space. Then the map M : X →
MaxSpec C (X ) is continuous when MaxSpec X is given the Zariski topology.
Proof. As above, it is enough to show that for all f ∈ C (X ), the preimage M−1 (V (f ))
is closed in X . Unpacking the deﬁnitions, we ﬁnd
M−1 (V (f )) = f −1 (0),
thus the preimage is the zero set of the continuous function f , hence closed.
Corollary 139. For a compact space X , the Zariski topology on M (X ) = MaxSpec C (X )
coincides with the initial topology.
Proof. By Theorem 137, we may instead compare the Zariski topology on X – i.e.,
the topology obtained by pulling back the Zariski topology on MaxSpec C (X ) via
M – with the given topology on X . But the proof of Proposition 138 shows that the
Zariski topology on X is precisely the Ztopology, i.e., the one in which the closed
subsets are the intersections of zero sets. But X is compact hence quasiTychonoﬀ,
so by Theorem 131 the Ztopology on X coincides with the given topology on
X.
Now let π : X → Y be a continuous map between compact spaces. There is an
induced map C (π ) : C (Y ) → C (X ): given g : Y → R, we pullback by π to get
g ◦ π : X → R. It is no problem to see that C (π ) is a homomorphism of rings. Now
let mx ∈ MaxSpec C (X ) be a maximal ideal and consider its pullback C (π )∗ (mx )
to an ideal of C (Y ): we ﬁnd
C (π )∗ (mx ) = {g : Y → R  g (π (x)) = 0} = mπ(x) .
Thus the pullback map carries maximal ideals to maximal ideals (recall this is certainly not true for all homomorphisms of rings!) and thus induces a map from X
to Y which is indeed nothing else than the given map π .
All in all we see that the functors C and MaxSpec give a duality between the
categories of compact spaces and rings of continuous Rvalued functions on compact spaces. In functional analysis this the ﬁrst step in an important circle of ideas
leading up to Gelfand duality for commutative Banach algebras.
What about the case of noncompact spaces X ?
Example: Let X be an inﬁnite discrete space, so C (X ) = RX is the ring of all
functions from X to R. Thus X is a noncompact Tychonoﬀ space. So it follows 88 PETE L. CLARK from our work so far that M gives a continuous injection from M to the quasicompact space MaxSpec C (X ). In fact M is an embedding: for any subset Y ⊂ X ,
let IY be the ideal of functions vanishing identically on Y . Then the restriction of
the closed subset V (I ) of MaxSpec C (X ) to M(X ) is precisely M(Y ), so M(Y ) is
closed in MaxSpec C (X ). Thus M(X ) is discrete as a subspace of MaxSpec C (X ),
and this implies that M is not surjective.
Theorem 140. For any topological space X , MaxSpec C (X ) endowed with the
Zariski topology is compact.
Theorem 141. Let X be any topological space, let M : X → MaxSpec C (X ), and
let XT = M(X ), viewed as a subspace of MaxSpec C (X ).
a) XT is a Tychonoﬀ space.
b) The map M : X → XT is the Tychonoﬀ completion of X : i.e., it is universal
for continuous maps from X to a Tychonoﬀ space.
c) The induced map C (M) : C (XT ) → C (X ) is an isomorphism of rings.
Proof. See [GJ76, §3.9].
Thus the ring of continuous functions on an arbitrary space X “sees” precisely its
Tychonoﬀ completion XT . Henceforth we restrict to Tychonoﬀ spaces.
Theorem 142. Let X be a Tychonoﬀ space. Then the map M : X → C (X ) is
nothing else than the StoneCech compactiﬁcation.
Proof. See [GJ76, §7.10].
Exercise 5.7: a) Show that C (X ) is an Rsubalgebra of RX .
b) Show that C (X ) is reduced: it contains no nonzero nilpotent elements.
Exercise 5.8: Show that C (X ) is connected in the algebraic sense – i.e., there
are no idempotents other than 0 and 1 – iﬀ the topological space X is connected.
Exercise 5.9: Show that there is an antitone Galois connection between 2X and
the set of ideals of C (X ), as follows:
S ⊂ X → IS = {f ∈ C (X )  f S ≡0 } and
I → VI = {x ∈ X  ∀f ∈ I, f (x) = 0}.
Exercise 5.10: Let X be a compact space.
a) Let p be a prime ideal of C (X ). Show that V (p) consists of a single point.
b) Deduce that a prime ideal p of C (X ) is closed in the sense of the Galois connection – i.e., p = IVp iﬀ p is maximal.
c) Deduce that each prime ideal p of C (X ) is contained in a unique maximal ideal.
Exercise 5.11: Let X = [0, 1] with the standard Euclidean topology. Let r0 be
(x
the ideal of all functions f ∈ C (X ) such that for all k ∈ N, limx→0+ fxk ) = 0.
Equivalently r0 is the ideal of all functions which are inﬁnitely diﬀerentiable at 0
and have identically zero Taylor series at zero.
a) Show that r0 is a radical ideal but not a prime ideal.
b) Show that the only maximal ideal containing r0 is m0 , the set of all functions
vanishing at 0.
c) Deduce that there exist ideals of C (X ) which are prime but not maximal. COMMUTATIVE ALGEBRA 89 Exercise 5.12: Let X be a Cseparated topological space.
a) Let S ⊂ X with #S > 1. Show that IS is not maximal.
b) Suppose X is Tychonoﬀ and S, T ⊂ X . Show that IS ⊂ IT ⇐⇒ T ⊂ S .
c) In particular if X is Tychonoﬀ, then for closed subsets S and T of X , IS =
IT ⇐⇒ S = T .
Exercise 5.13: Let φ : X → Y be a continuous function between topological spaces.
a) Show that φ induces a ring homomorphism C (φ) : C (Y ) → C (X ) by g ∈
C (Y ) → φ∗ g = g ◦ φ.
b) Suppose Y is normal, X is a closed subspace of X and φ : X → Y is the inclusion
map. Show that C (φ) is surjective.
Exercise 5.14: Let X be a normal topological space. Show that the closure operator on subsets of X given by the Galois connection coincides with the topological
closure operator on X .
1
Exercise 5.15: Let X be the subspace { n }n∈Z+ ∪{0} of R, and let m be the maximal
ideal of all functions vanishing at 0. Fill in the details of the following outline of
a proof that m is not ﬁnitely generated.23 Assume otherwise: m = ⟨a1 , . . . , an ⟩.
g2 (
Then for every element g ∈ m, limx→0 a1 (x)+...x)an (x) = 0. (In particular, there
+
exists δ > 0 such that the denominator is strictly positive on (0, δ ).) Now choose
g ∈ m so as to get a contradiction. Exercise 5.16: Let X be a normal space, and let x ∈ X .
a) Show that the following are equivalent:
(i) The ideal Ix is ﬁnitely generated.
(ii) The ideal Ix is principal.
(iii) The point x is isolated in X (i.e., {x} is open).
b) Suppose X is compact. Show that the following are equivalent:
(i) C (X ) is a Noetherian ring.
(ii) C (X ) is ﬁnitedimensional as an Rvector space.
(iii) X is ﬁnite.
Exercise 5.17: Show that if we worked throughout with rings C (X, C) of continuous
Cvalued functions, then all of the above results continue to hold.
Exercise 5.18: Suppose that we looked at rings of continuous functions from a topological space X to Qp . To what extent to the results of the section continue to hold?
Exercise 5.19: Let X be a compact smooth manifold and consider the ring C∞ (X )
of smooth functions f : X → R.
a)* Show that for x ∈ X , the ideal mx of all functions vanishing at x is maximal
and ﬁnitely generated.
b) Note that the phenomenon of part a) is in contrast to the case of maximal ideals
in the ring C ([0, 1]), say. However, I believe that with this sole exception, all of the
23Or, if you like, give your own proof that m is not ﬁnitely generated. 90 PETE L. CLARK results of this section hold for the rings C∞ (X ) just as for the rings C (X ). Try it
and see.
5.3. Rings of holomorphic functions.
As we saw in the previous section, one of the characteristic properties of the ring of
continuous functions on a normal space (or even smooth functions on a manifold)
is that it is typically very far from being an integral domain. A remedy for this is
to consider more “rigid” collections of functions.
Let U be an open subset of the complex plane C, and let Hol(U ) be the set of
holomorphic functions f : U → C. (Recall that a holomorphic function on U is one
for which the complex derivative f ′ (z ) exists for each z ∈ U . Equivalently, for each
z ∈ U f admits a power series development with positive radius of convergence.)
It is immediate that Hol(U ) is a subring of the ring CU of all Cvalued functions
on U .
Proposition 143. For a nonempty open subset U of C, the following are equivalent:
(i) U is connected.
(ii) Hol(U ) is a domain.
Proof. (i) =⇒ (ii): For any f ∈ C (U, C) let Z (f ) = {z ∈ U  f (z ) = 0} be the
zero set of f . Since f is continuous, Z (f ) is a closed subset of U . If f is moreover
holomorphic, then Z (f ) has no accumulation point in U , i.e., f ̸= 0 =⇒ Z (f ) is
discrete – 24 in particular Z (f ) is countable. Moreover, for any f, g ∈ C (U, C) we
have Z (f g ) = Z (f ) ∪ Z (g ), so if f, g ∈ Hol(U )• then Z (f g ) is at most countable,
whereas U is uncountable, so f g ̸= 0.
(ii) =⇒ (i): we argue by contrapositive. If U is not connected, it is of the form V1 ∪
V2 where V1 and V2 are disjoint open subsets. Let χi be the characteristic function
of Vi for i = 1, 2. Then each χi is locally constant on U – hence holomorphic, and
nonzero, but χ1 χ2 = 0.
Recall that in complex function theory it is common to call a nonempty open subset
U of C a domain. In this language, Proposition 143 simply asserts that U is a
domain iﬀ Hol(U ) is a domain. Henceforth we assume that U is a domain.
For every z ∈ U there is a function ordz : Hol(U )• → N, the order of vanishing
∑∞
of f at z. Precisely, we exand f into a power series at z : f (ζ ) = n=0 an (ζ − z )n
and put ordz (f ) to the least n for which an ̸= 0. Compiling all these together
we may associate to each f ∈ Hol(U )• its total order Ord(f ) : U → N given by
Ord(f )(z ) = ordz (f ).
Consider the set NU of all functions from U to N. For O ∈ NU , we deﬁne the
support of O to be the set of z ∈ U such that O(z ) > 0.
Recall that a meromorphic function on U is a function which is holomorphic on U
except for isolated ﬁnite order singularities. More precisely, a meromorphic function
is a function which is holomorphic on U \ Z for some discrete closed subset Z of U
24Recall that this is proved by considering the Taylor series development of f about any
accumulation point. COMMUTATIVE ALGEBRA 91 and such that for all z0 ∈ Z , there exists n ∈ Z+ such that (z − z0 )n f (z ) extends to
a holomorphic function on a neighborhood of z . If the least n as above is positive,
we say that f has a pole at z0 , and we employ the convention that f (z0 ) = ∞. Let
Mer(U ) be the set of all meromorphic functions on U ; it is a ring under pointwise
addition and multiplication, under the conventions that for all z ∈ C,
z + ∞ = ∞ + ∞ = z · ∞ = ∞ · ∞ = ∞.
Theorem 144. Let U be a domain in the complex plane.
a) (Weierstrass) For each O ∈ NU with closed, discrete support, there exists f ∈
Hol(U )• with Ord(f ) = O.
b) (Weierstrass + MittagLeﬄer) Let Z ⊂ U be a closed subset without limit points.
To each z ∈ Z we associate a natural number nz and for all 0 ≤ k ≤ nz , a complex
number wz,k . Then there exists f ∈ Hol(U ) such that for all z ∈ Z and 0 ≤ k ≤ nz ,
f (k) (z ) = k !wz,k .
Proof. Part a) is of the two main results in Weierstrass’ Factorization Theory: see
e.g. [Ru87, Thm. 15.11]. Part b) is proved by combining part a) with MittagLeﬄer’s famous result on the existence of meromorphic functions with prescribed
principal parts: see e.g. [Ru87, Thm. 15.13].
Corollary 145. The ring Mer(U ) of meromorphic functions on U is a ﬁeld, and
indeed is the ﬁeld of fractions of Hol(U ).
Exercise 5.20: Prove Proposition 145.
Exercise 5.21: Fix z0 ∈ U . For f ∈ Mer(U ), choose n ∈ N such that (z − z0 )n f is
holomorphic at z0 , and put ordz0 (f ) = ordz0 ((z − z0 )n f ) − n.
a) Show that this gives a welldeﬁned function ordz0 : Mer(U )• → Z (i.e., independent of the choice of n in the deﬁnition).
b) Show that for all f, g ∈ Mer(U )× , ordz0 (f g ) = ordz0 (f ) + ordz0 (g ).
c) We formally extend ordz0 to a function from Mer(U ) to Z ∪ {∞} by setting
ordz0 (0) = ∞. Show that, under the convention that ∞ + n = ∞ + ∞ = ∞, we
have for all f, g ∈ Mer(U ) that ordz0 (f + g ) ≥ min ordz0 (f ), ordz0 (g ).
d) Show that if ordz0 (f ) ̸= ordz0 (g ) then ordz0 (f + g ) = min ordz0 f, ordz0 (g ).
Similarly we may extend Ord to a function from Mer(U )• to ZU .
Lemma 146. For f, g ∈ Hol(U )• , TFAE:
(i) Ord(f ) = Ord(g ).
(ii) f = ug for u ∈ Hol(U )× .
(iii) (f ) = (g ).
Proof. (ii) ⇐⇒ (iii) for elements of any integral domain.
(ii) =⇒ (i) is easy and left to the reader.
(i) =⇒ (ii): The meromorphic function f has identically zero order, hence is
g
nowhere vanishing and is thus a unit u in Hol(U ).
Theorem 147. (Helmer [Hel40]) For a domain U in the complex plane, every
ﬁnitely generated ideal of Hol(U ) is principal. More precisely, for any f1 , . . . , fn ∈
Hol(U )• , there exists f ∈ Hol(U ) such that Ord(f ) = mini Ord(fi ), unique up to
associates, and then ⟨f1 , . . . , fn ⟩ = ⟨f ⟩. 92 PETE L. CLARK Proof. Step 1: Suppose that f1 , f2 ∈ Hol(U )• do not simultaneously vanish at any
point of U . We claim that ⟨f1 , f2 ⟩ = Hol(R).
proof of claim Let Z be the zero set of f1 , so that for all z ∈ Z , f2 (z ) ̸= 0. By
Theorem 144b) there exists g2 ∈ Hol(U ) such that for all z ∈ Z , ordz (1 − g2 f2 ) ≥
g2
ordz (f1 ). Thus Ord(1 − g2 f2 ) ≥ Ord(f1 ), so that g1 := 1−f1 f2 ∈ Hol(U ) and thus
f1 g1 + f2 g2 = 1.
Step 2: Now let f1 , f2 ∈ Hol(U )• be arbitrary. By Theorem 144a), there exists
f ∈ Hol(U ) with Ord(f ) = min Ord(f1 ), Ord(f2 ). For i = 1, 2, put gi = fi . Then g1
f
and g2 are holomorphic and without a common zero, so by Step 1 ⟨g1 , g2 ⟩ = Hol(U ).
Multiplying through by f gives ⟨f1 , f2 ⟩ = ⟨f ⟩.
Step 3: An easy induction argument shows that in a ring R in which every ideal
of the form ⟨x1 , x2 ⟩ is principal, every nitely generated ideal is principal. By
Step 2, this applies in particular to Hol(U ). Moreover, it is easy to see that if
the ideal ⟨f1 , . . . , fn ⟩ is generated by any single element f , then we must have
Ord f = min Ord fi .
Exercise 5.22: Explain carefully why in Step 1 of the above proof, Theorem 144b)
implies the existence of g2 .
Of course the most familiar class of domains in which every ﬁnitely generated ideal
is principal are those domains in which every ideal is principal: PIDs! But as the
reader has probably already suspected, Hol(U ) is not a PID.
One way to see this is to show that Hol(U ) is not even a UFD. Remarkably, this is
an immediate consequence of the Weierstrass Factorization Theory, which succeeds
in decomposing every holomorphic function into a product of prime elements! The
catch is that most holomorphic functions require inﬁnite products, a phenomenon
which is not countenanced in the algebraic theory of factorization.
Exercise 5.23: Let f ∈ Hol(U )• .
a) Show that f is an irreducible element of Hol(U ) – i.e., if f = g1 g2 then exactly
one of g1 , g2 is a unit – iﬀ it has exactly one simple zero.
b) Suppose f is irreducible. Show that Hol(U )/(f ) = C. In particular, (f ) is a
prime ideal.
c) Show that f admits a (ﬁnite!) factorization into irreducible elements iﬀ it has
only ﬁnitely many zeros. Conclude that Hol(U ) is not a UFD.
Exercise 5.24: Extract from the previous exercise an explicit ideal of Hol(C) which
is not ﬁnitely generated.
Exercise 5.25*: Show that all of the results of this section extend to the ring of
holomorphic functions on a noncompact Riemann surface.
Exercise 5.26*: Investigate the extent to which the results of this section continue
to hold for Stein manifolds. (Step 1: learn the deﬁnition of a Stein manifold!)
5.4. Polynomial rings. COMMUTATIVE ALGEBRA 93 Let R be a ring (possibly noncommutative, but – as ever – with identity). Then
R[t] denotes the ring of univariate polynomials with Rcoeﬃcients.
We assume the reader knows what this means in at least an informal sense: an
element of R will be an expression of the form an tn + . . . + a1 t + a0 , where n is some
nonnegative integer and an , . . . , a0 are in R. The degree of a polynomial is the
supremum over all numbers n such that an ̸= 0. We say “supremum” rather than
“maximum” as an attempt to justify the convention that the degree of the 0 polynomial should be −∞ (for that is the supremum of the empty set). A polynomial
of degree 0 is called constant, and we can view R as a subset of R[t] by mapping a ∈ R to the constant polynomial a. As an abelian group, R[t] is canonically
⊕∞
∑
n
isomorphic to
n=0 R, the isomorphism being given by
n an t → (a0 , a1 , . . .).
(The key point here is that on both sides we have an = 0 for all suﬃciently large
n.) Multiplication of polynomials is obtained by applying the relations t0 = 1,
ti+j = ti tj at = ta for all a ∈ R, and distributivity, i.e.,
∑
(an tn + . . . + an t1 + a0 ) · (bm tm + . . . + b1 t + b0 ) =
ai bj ti+j .
0≤i≤n, 0≤j ≤m For any P ∈ R[t], the identity 1 ∈ R has the property 1 · P = P · 1 = 1.
Unfortunately there are some minor annoyances of rigor in the previous description. The ﬁrst one – which a suﬃciently experienced reader will immediately either
dismiss as silly or know how to correct – is that it is not settheoretically correct :
technically speaking, we need to say what R[t] is as a set and this involves saying
what t “really is.” It is common in abstract algebra to refer to t is an indeterminate, a practice which is remarkably useful despite being formally meaningless:
essentially it means “Don’t worry about what t is; it can be anything which is not
an element of R. All we need to know about t is encapsulated in the multiplication
rules at = ta, t0 = 1, ti tj = ti+j .” In other words, t is what in the uncomplicated
days of high school algebra was referred to as a variable.
If someone insists that R[t] be some particular set – a rather unenlighened attitude that we will further combat later on – then the solution has already been
⊕∞
given: we can take R[t] =
n=0 R. (It is fair to assume that we already know
what direct sums of abelian groups “really are”, but in the next section we will
give a particular construction which is in fact rather useful.) This disposes of the
settheoretic objections.
Not to be laughed away completely is the following point: we said R[t] was a
ring, but how do we know this? We did explain the group structure, deﬁned a multiplication operation, and identiﬁed a multiplicative identity. It remains to verify
the distributivity of multiplication over addition (special cases of which motivated
our deﬁnition of multiplication, but nevertheless needs to be checked in general)
and also the associativity of multiplication.
Neither of these properties are at all diﬃcult to verify. In fact:
Exercise 5.27: a) Show that R[t] is a ring.
b) Show that R[t] is commutative iﬀ R is commutative. 94 PETE L. CLARK Let us now attempt a “conceptual proof” of the associativity of polynomial multiplication. For this we shall assume that R is commutative – this is the only case
we will be exploring further anyway. Then we can, as the P (t) notation suggests,
view an element of R[t] as a function from R to R. Namely, we just plug in values:
a ∈ R → P (a) ∈ R.
To be clear about things, let us denote this associated function from R to R by
P . As we saw above, the set of all functions RR from R to R forms a commutative ring under pointwise addition and multiplication: (f + g )(a) := f (a) + g (a),
(f g )(a) := f (a) · g (a). In particular, it really is obvious that the multiplication of
functions is associative. Let P be the subset of RR of functions of the form P for
some P ∈ R[t]. More concretely, we are mapping the constant elements of R[t] to
constant functions and mapping t to the identity function. This makes it clear that
P is a subring of RR : in fact it is the subring of RR generated by the constant
functions and the identity function.
So why don’t we just deﬁne R[t] to be P , i.e., identify a polynomial with its associated function?
The problem is that the map R[t] → P need not be an injection. Indeed, if R
is ﬁnite (but not the zero ring), P is a subring of the ﬁnite ring RR so is obviously
ﬁnite, whereas R[t] is just as obviously inﬁnite. If R is a domain this turns out to
be the only restriction.
Proposition 148. Let R be an integral domain.
a) Suppose that R is inﬁnite. Then the canonical mapping R[t] → P is a bijection.
b) Suppose that R is ﬁnite, say of order q , and is therefore a ﬁeld. Then the kernel
of the canonical mapping R[t] → P is the principal ideal generated by tq − t.
We leave the proof as a (nontrivial) exercise for the interested reader.
Exercise 5.28: Exhibit an inﬁnite commutative ring R for which the map R[t] → P
is not injective. (Suggestion: ﬁnd an inﬁnite ring all of whose elements x satisfy
x2 = x.)
Exercise 5.29: Show that the map R[t] → P is a homomorphism of rings.
So if we restrict to inﬁnite integral domains, the map R[t] → P is an isomorphism of rings. Thus we see, after the fact, that we could have deﬁned the ring
structure in terms of pointwise multiplication.
5.5. Semigroup algebras.
A semigroup M is a set equipped with a single binary operation ·, which is required (only!) to be associative. A monoid is a semigroup with a twosided identity.
Exercise 5.30: Show that a semigroup has at most one twosided identity, so it
is unambiguous to speak of “the” identity element in a monoid. We will denote it
by e (so as not to favor either addditive or multiplicative notation). COMMUTATIVE ALGEBRA 95 Example: Let (R, +, ·) be an algebra. Then (R, ·) is a semigroup. If R is a ring
(i.e., has an identity 1) then (R, ·) is a monoid, with identity element 1.
Example: Any group is a monoid. In fact a group is precisely a monoid in which
each element has a twosided inverse.
Example: The structure (N, +) of natural numbers under addition is a monoid;
the identity element is 0.
Example: The structure (Z+ , ·) of positive integers under multiplication is a monoid;
the identity element is 1.
Let M and N be two semigroups. Then the Cartesian product M × N becomes
a semigroup in an obvious way: (m1 , n1 ) · (m2 , n2 ) := (m1 · m2 , n1 · n2 ). If M
and N are monoids with identity elements eM and eN , then M × N is a monoid,
with identity element (eM , eN ). Exactly the same discussion holds⊕ any ﬁnite
for
n
set M1 , . . . , MN of semigroups: we can form the direct sum M =
i=1 Mi , i.e.,
the Cartesian product of sets with componentwise operations; if all the Mi ’s are
monoids, so is M .
If we instead have an inﬁnite family {Mi }i∈I of semigroups indexed by a set I , we
∏
can deﬁne a semigroup structure on the Cartesian product i∈I Mi in the obvious
way, and if each Mi is a monoid with identity ei , then the product semigroup is a
monoid with identity (ei )i∈I . If each Mi is a monoid, we can also deﬁne the direct
⊕
∏
sum i∈I Mi , which is the subset of the direct product i∈I Mi consisting of all
I tuples (mi ∈ Mi )i∈I such that mi = ei for all but ﬁnitely many i. Then we have
⊕
∏
that i∈I Mi is a submonoid of the direct product monoid i∈I Mi .
If M and N are semigroups, then a map f : M → N is a homomorphism of
semigroups if f (m1 · m2 ) = f (m1 ) · f (m2 ) for all m1 , m2 ∈ M . If M and N are
monoids, a homomorphism of monoids is a homomorphism of semigroups such that
moreover f (eM ) = eN . A homomorphism f : M → N of semigroups (resp. of
monoids) is an isomorphism iﬀ there exists a homomorphism of semigroups (resp.
monoids) g : N → M such that g ◦ f = IdM , f ◦ g = IdN .
Exercise 5.31: a) Exhibit monoids M and N and a homomorphism of semigroups
f : M → N which is not a homomorphism of monoids.
b) Show that a homomorphism of semigroups f : M → N is an isomorphism iﬀ it
is bijective. Same for monoids.
Exercise 5.32: Show that the monoid (Z+ , ·) of positive integers under multipli⊕∞
cation is isomorphic to i=1 (N, +), i.e., the direct sum of inﬁnitely many copies
of the natural numbers under addition. (Hint: a more natural indexing set for the
direct sum is the set of all prime numbers.)
Now let R be an algebra and M be a semigroup. We suppose ﬁrst that M is
ﬁnite. Denote by R[M ] the set of all functions f : M → R. 96 PETE L. CLARK As we saw, using the operations of pointwise addition and multiplication endow
this set with the structure of an associative algebra (which has an identity iﬀ M
does). We are going to keep the pointwise addition but take a diﬀerent binary
operation ∗ : R[M ] × R[M ] → R[M ].
Namely, for f, g ∈ R[M ], we deﬁne the convolution product f ∗ g as follows:
∑
(f ∗ g )(m) :=
f (a)g (b).
(a,b)∈M 2  ab=m In other words, the sum extends over all ordered pairs (a, b) of elements of M whose
product (in M , of course), is m.
Proposition 149. Let R be an associative algebra and M a ﬁnite semigroup. The
structure (R[M ], +, ∗) whose underlying set is the set of all functions from M to
R, and endowed with the binary operations of pointwise additition and convolution
product, is an associative algebra. If R is a ring and M is a monoid with identity
e, then R[M ] is a ring with multiplicative identity the function I which takes eM
to 1R and every other element of M to 0R .
Proof. First, suppose that R is a ring and M is a monoid, then for any f ∈ R[M ]
and m ∈ M , we have
∑
(f ∗I )(m) =
f (a)I (b) = f (m)I (1) = f (m) = I (1)f (m) = . . . = (I ∗f )(m).
(a,b)∈M 2  ab=m We still need to check the associativity of the convolution product and the distributivity of convolution over addition. We leave the latter to the reader but check the
former: if f, g, h ∈ R[M ], then
∑
∑∑
((f ∗ g ) ∗ h)(m) =
(f ∗ g )(x)h(c) =
f (a)g (b)h(c)
xc=m =
= ∑∑
ay =m bc=y ∑ xc=m ab=x f (a)g (b)h(c) abc=m f (a)g (b)h(c) = ∑ f (a)(g ∗ h)(y ) = (f ∗ (g ∗ h))(m). ay =m A special case of this construction which is important in the representation theory
of ﬁnite groups is the ring k [G], where k is a ﬁeld and G is a ﬁnite group.
Now suppose that M is an inﬁnite semigroup. Unless we have some sort of extra structure on R which allows us to deal with convergence of sums – and, in
this level of generality, we do not – the above deﬁnition of the convolution product
f ∗ g is problematic because the sum might be inﬁnite. For instance, if M = G
is any group, then our previous deﬁnition of (f ∗ g )(m) would come out to be
∑
−1
m), which is, if G is inﬁnite, an inﬁnite sum.
x∈G f (x)g (x
Our task therefore is to modify the construction of the convolution product so
as to give a meaningful answer when the semigroup M is inﬁnite, but in such a way
that agrees with the previous deﬁnition for ﬁnite M . COMMUTATIVE ALGEBRA 97 Taking our cue from the inﬁnite direct sum, we restrict our domain: deﬁne R[M ] to
be subset of all functions f : M → R such that f (m) = 0 except for ﬁnitely many
m (or, for short, ﬁnitely nonzero functions). Restricting to such functions,
∑
(f ∗ g )(m) :=
f (a)g (b)
ab=m makes sense: although the sum is apparently inﬁnite, all but ﬁnitely terms are zero.
Proposition 150. Let R be an associative algebra and M a semigroup. The structure (R[M ], +, ∗) whose underlying set is the set of all ﬁnitely nonzero functions
from M to R, and endowed with the binary operations of pointwise additition and
convolution product, is an associative algebra. If R is a ring and M is a monoid
with identity element e, then R[M ] is a ring with multiplicative identity the function
I which takes eM to 1R and every other element of M to 0R .
Exercise 5.33: Prove Proposition 150. More precisely, verify that the proof of
Proposition 149 goes through completely unchanged.
Note that as an abelian group, R[M ] is naturally isomorphic to the direct sum
⊕
m∈M R, i.e., of copies of R indexed by M . One can therefore equally well view
∑
an element R[M ] as a formal ﬁnite expressions of the form m∈M am m, where
am ∈ R and all but ﬁnitely many are 0. Written in this form, there is a natural
way to deﬁne the product
(
)(
)
∑
∑
am m
bm m
m∈M m∈M of two elements f and g of R[M ]: namely we apply distributivity, use the multiplication law in R to multiply the am ’s and the bm ’s, use the operation in M to
multiply the elements of M , and then ﬁnally use the addition law in R to rewrite
∑
the expression in the form m cm m. But a moment’s thought shows that cm is
nothing else than (f ∗ g )(m). On the one hand, this makes the convolution product
look very natural. Conversely, it makes clear:
The polynomial ring R[t] is canonically isomorphic to the monoid∑
ring R[N]. Indeed, the explict isomorphism is given by sending a polynomial n an tn to the
function n → an .
This gives a new proof of the associativity of the product in the polynomial ring
R[t]. We leave it to the reader to decide whether this proof is any easier than direct
veriﬁcation.. Rather the merit is that this associativity computation has been done
once and for all in a very general context.
As an aside, let me point out something very curious: in searching for a slick
proof of associativity of multiplication in the polynomial ring R[t], we attempted
to show that the multiplication was just multiplication of the associated functions
f : R → R. As we saw, this works in many but not all cases (because the homomorphism from R[t] to the ring of functions RR is not always surjective). Then
we observed that the associativity of multiplication of polynomials is a special case
of associativity of the product in a semigroup ring. What is strange is that the 98 PETE L. CLARK elements of this semigroup ring R[N] are themselves deﬁned as functions, but functions from N to R, and the product is not the most obvious (pointwise) one –
which has nothing to do with the semigroup structure on N – but rather a “funny”
convolution product. Suitably mathematically urbane readers will, upon seeing a
homomorphism from an abelian group (here R[t]) to another abelian group (here
P ) which converts a “convolution product” on the ﬁrst group to a “pointwise product” on the second group, be tempted to view the mapping R[t] → P as some sort
of Fourier transform. If I understood this phenomenon more completely myself,
I might be more willing to digress to explain it (but I don’t, so I won’t).
The semigroup algebra construction can be used to deﬁne several generalizations
of the polynomial ring R[t].
Exercise 5.34: For any ring R, identify the monoid ring R[Z] with the ring R[t, t−1 ]
of Laurent polynomials.
⊕
First, let T = {ti } be a set. Let F A(T ) :=
i∈T (N, +) be the direct sum of a
number of copies of (N, +) indexed by T . Let R be a ring, and consider the monoid
ring R[F A(T )]. Let us write the composition law in F A(T ) multiplicatively; moreover, viewing an arbitrary element I of F A(T ) as a ﬁnitely nonzero function from T
∏
to N, we use the notation tI for t∈T tI (t) . Then an arbitrary element of R[F A(T )]
∑n
is a ﬁnite sum of the form k=1 rk tIk , where I1 , . . . , Ik are elements of F A(t). This
representation of the elements should make clear that we can view R[F A(T )] as
a polynomial ring in the indeterminates t ∈ T : we use the alternate notation R[{ti }].
Let us go back to the monoid ring R[N], whose elements are ﬁnitely nonzero functions f : R → N. Notice that in this case the precaution of restricting ﬁnitely
nonzero functions is not necessary: the monoid (N, +), although inﬁnite, has the
property that for any m ∈ N, the set of all x, y ∈ N such that x + y = m is ﬁnite
(indeed, of cardinality m + 1). Let us call an arbitrary monoid M divisorﬁnite
if for each m in M , the set {(x, y ) ∈ M 2  xy = m} is ﬁnite.
Exercise 5.35: a) For any set T , the monoid F A(T ) =
ﬁnite.
b) A group is divisorﬁnite iﬀ it is ﬁnite. ⊕ t∈T (N, +) is divisor For a divisorﬁnite monoid M , and any ring R, we may deﬁne the big monoid
ring R[[M ]] to be the collection of all functions M → R, with pointwise addition
and convolution product.
For example, if M = (N, +), then writing M multiplicatively with n ∈ N → tn for
some formal generator t, an element of the ring R[[M ]] is an inﬁnite formal sum
∑
n
n∈N rn t . Such sums are added coordinatewise and multiplied by distributivity:
( ∑ n∈N rn tn )( ∑ n∈N sn tn ) = r0 s0 + (r0 s1 + r1 s0 )t + . . . + ( n
∑ rk sn−k )tn + . . . . k=0 This ring is denoted by R[[t]] and called the formal power series ring over R. COMMUTATIVE ALGEBRA 99 Exercise 5.36: Using Exercise 5.35, deﬁne, for any set T = {ti } and any ring
R, a formal power series ring R[[{ti }]].
Here is yet another variation on the construction: suppose M is a commutative,
cancellative divisorﬁnite monoid endowed with a total order relation ≤. (Example:
(N, +) or F A(T ) for any T .) There is then a group completion G(M ) together with
an injective homomorphism of monoids M → G(M ). If M is ﬁnite and cancellative,
it is already a group. If M is inﬁnite, then so is G(M ), so it cannot be divisorﬁnite.
Nevertheless, the ordering ≤ extends uniquely to an ordering on G(M ), and we can
deﬁne a ring R((G(M )) whose elements are the functions from f : G(M ) → R
such that {x ∈ G(M )  x < 0, f (x) ̸= 0} is ﬁnite, i.e., f is ﬁnitely nonzero on the
negative values of G(M ).
Exercise 5.37: a) Show that under the above hypotheses, the convolution product on R((G(M )) is welldeﬁned, and endows R((G(M )) with the structure of a
ring.
b) When M = (N, +), identify R((M )) as R((t)), the ring of formal ﬁnitetailed
Laurent series with coeﬃcients in R. Give a multivariable analogue of this by
taking M = F A(T ) for arbitrary T .
Exercise 5.38: Let R be a possibly noncommutative ring. Give a rigorous deﬁnition of the ring R⟨t1 , t2 ⟩ of “noncommutative polynomials” – each ti commutes
with each element of R, but t1 and t2 do not commute – as an example of a small
monoid ring R[M ] for a suitable monoid M . Same question but with an arbitrary
set T = {ti } of noncommuting indeterminates.
The universal property of semigroup rings: Fix a commutative ring R. Let B
be a commutative Ralgebra and M a commutative monoid. Let f : R[M ] → B be
an Ralgebra homomorphism. Consider f restricted to M ; it is a homomorphism
of monoids M → (B, ·). Thus we have deﬁned a mapping
HomRalg (R[M ], B ) → HomMonoid (M, (B, ·)).
Interestingly, this map has an inverse. If g : M → B is any homomorphism satisfying g (0) = 0, g (m1 + m2 ) = g∑ 1 ) + g (m2 ), then g extends to a unique Ralgebra
(m
∑
homomorphism R[M ] → B :
m∈M rm m →
m rm g (m). The uniqueness of the
extension is immediate, and that the extended map is indeed an Ralgebra homomorphism can be checked directly (please do so).
In more categorical language, this canonical bijection shows that the functor M →
R[M ] is the left adjoint to the forgetful functor (S, +, ·) → (S, ·) from Ralgebras
to commutative monoids. Yet further terminology would express this by saying
that R[M ] is a “free object” of a certain type.
Theorem 151. (Universal property of polynomial rings) Let T = {ti } be a set of
indeterminates. Let R be a commutative ring, and S an Ralgebra. Then each map
of sets T → S extends to a unique Ralgebra homomorphism R[T ] → S .
Proof: By the previous result, each monoid map from the free commutative monoid
⊕
t∈T Z to S extends to a unique Ralgebra homomorhpism. So what is needed is
the fact that every set map T → M to a commutative monoid extends uniquely 100 PETE L. CLARK ⊕
to a homomorphism t∈T Z → M (in other words, we pass from the category of
sets to the category of commutative Ralgebras by passing through the category
of commutative monoids, taking the free commutative monoid associated to a set
and then the free Ralgebra associated to the monoid). As before, the uniqueness
of the extension is easy to verify.
Exercise 5.39: a) Formulate analogous universal properties for Laurent polynomial rings, and noncommutative polynomial rings.
b) Suppose M is a divisorﬁnite monoid. Is there an analogous extension property
for the big monoid ring R[[M ]]?
This result is of basic importance in the study of Ralgebras. For instance, let
S be an Ralgebra. A generating set for S , as an Ralgebra, consists of a subset T
of S such that the least Rsubalgebra of S containing T is S itself. This deﬁnition
is not very concrete. Fortunately, it is equivalent to the following:
Theorem 152. Let R be a commutative ring, S a commutative Ralgebra, and T
a subset of S . TFAE:
(i) T generates S as an Ralgebra.
(ii) The canonical homomorphism of Ralgebras R[T ] → S – i.e., the unique one
sending t → t – is a surjection.
Exercise 5.40: Prove Theorem 152.
In particular, a commutative Ralgebra S is ﬁnitely generated iﬀ it is a quotient
ring of some polynomial ring R[t1 , . . . , tn ].
Another application is that every commutative ring whatsoever is a quotient of
a polynomial ring (possibly in inﬁnitely many indeterminates) over Z. Indeed, for
a ring R, there is an obvious surjective homomorphism from the polynomial ring
Z[R] – here R is being viewed as a set of indeterminates – to R, namely the one
carrying r → r.
A ring R is said to be absolutely ﬁnitely generated if it is ﬁnitely generated as
a Zalgebra; equivalently, there exists an n ∈ N and an ideal I in Z[t1 , . . . , tn ] such
that Z[t1 , . . . , tn ] is isomorphic to R.
Exercise 5.41: a) Show that every ﬁnitely generated ring has ﬁnite or countably
inﬁnite cardinality.
b) Find all ﬁelds which are ﬁnitely generated as rings. (N.B.: In ﬁeld there is another notion of absolute ﬁnite generation for a ﬁeld. This a much weaker notation:
e.g. Q(x) is absolutely ﬁnitely generated as a ﬁeld but not as a ring.)
6. Swan’s Theorem
We now digress to discuss an important theorem of R.G. Swan on projective modules over rings of continuous functions.
Throughout this section K denotes either the ﬁeld R or the ﬁeld C, each endowed
with their standard Euclidean topology. For a topological space X , the set C (X ) COMMUTATIVE ALGEBRA 101 of all continuous functions f : X → K forms a commutative ring under pointwise
addition and multiplication.
6.1. Introduction to (topological) vector bundles.
Recall25 the notion of a Kvector bundle over a topological space X . This is given
by a topological space E (the “total space”), a surjective continuous map π : E → X
and on each ﬁber Ex := π −1 (x) the structure of a ﬁnitedimensional K vector space
satisfying the following local triviality property: for each x ∈ X , there exists an
open neighborhood U containing x and a homeomorphism f : π −1 U → U × K n
such that for all y ∈ U f carries the ﬁber Ey over y to {y } × K n and induces on
these ﬁbers an isomorphism of K vector spaces. (Such an isomorphism is called
a local trivialization at x.) As a matter of terminology we often speak of “the
vector bundle E on X ” although this omits mention of some of the structure.
On any K vector bundle E over X we have a rank function r : X → N, namely
we deﬁne r(x) to be the dimension of the ﬁber Ex . We say that E is a rank n
vector bundle if the rank function is constantly equal to n. The existence of local
trivializations implies that the rank function is locally constant – or equivalently,
continuous when N is given the discrete topology, so if the base space X is connected the rank function is constant.
As a basic and important example, for any n ∈ N we have the trivial rank n
vector bundle on X , with total space X × K n and such that π is just projection
onto the ﬁrst factor.
If π : E → X and π ′ : E ′ → X are two vector bundles over X , a morphism
of vector bundles f : E → E ′ is a continuous map of topological spaces from E to
E ′ over X in the sense that π = π ′ ◦ f – equivalently f sends the ﬁber Ex to the
′
ﬁber Ex – and induces a K linear map on each ﬁber. In this way we get a category
Vec(X ) of K vector bundles on X . If we restrict only to rank n vector bundles and
morphisms between them we get a subcategory Vecn (X ). A vector bundle E on X
is said to be trivial (or, for emphasis, “globally trivial”) if it is isomorphic to the
trivial rank n vector bundle for some n.
Many of the usual linear algebraic operations on vector spaces extend immediately
to vector bundles. Most importantly of all, if E and E ′ are two vector bundles on
X , we can deﬁne a direct sum E ⊕ E ′ , whose deﬁnining property is that its ﬁber over
each point x ∈ X is isomorphic to Ex ⊕ Ex′ . This not being a topology/geometry
course, we would like to evade the precise construction, but here is the idea: it is
obvious how to deﬁne the direct sum of trivial bundles. So in the general case, we
deﬁne the direct sum by ﬁrst restricting to a covering family {Ui }i∈I of simultaneous local trivializations of E and E ′ and then glue together these vector bundles
over the Ui ’s. In a similar way one can deﬁne the tensor product E ⊗ E ′ and the
dual bundle E ∨ .
For our purposes though the direct sum construction is the most important. It
25from a previous life, if necessary 102 PETE L. CLARK gives Vec(X ) the structure of an additive category: in addition to the existence
of direct sums, this means that each of the sets Hom(E, E ′ ) of morphisms from E
to E ′ form a commutative group. (In fact Hom(E, E ′ ) naturally has the structure
of a K vector space.) Decategorifying, the set of all isomorphism classes of vector
bundles on X naturally forms a commutative monoid under direct sum (the identity is the trivial vector bundle X → X where each one point ﬁber is identiﬁed –
uniquely! – with the zero vector space). The Grothendieck group of this monoid is
K (X ): this is the beginning of topological Ktheory.
6.2. Swan’s Theorem.
But we digress from our digression. A (global) section of a vector bundle π :
E → X is indeed a continuous section σ of the map π , i.e., a continuous map
σ : X → E such that π ◦ σ = 1X . The collection of all sections to E will be denoted
Γ(E ). Again this is a commutative group and indeed a K vector space, since we
can add two sections and scale by elements of K .
But in fact more is true. The global sections form a module over the ring
C (X ) of continuous K valued functions, in a very natural way: given a section
σ : X → E and f : X → K , we simply deﬁne f σ : X → E by x → f (x)σ (x). Thus
Γ : E → Γ(E ) gives a map from vector bundles over X to C (X )modules.
In fancier language, Γ gives an additive functor from the category of vector
bundles on X to the category of C (X )modules; let us call it the global section
functor. (Indeed, if we have a section σ : E → X of E and a morphism of vector
bundles f : E → E ′ , f (σ ) = f ◦ σ is a section of E ′ . No big deal!)
Theorem 153. (Swan [Sw62]) Let X be a compact space. Then the global section
functor Γ gives an equivalence of categories from Vec(X ) to the category of ﬁnitely
generated projective C (X )modules.
In other words, at least for this very topologically inﬂuenced class of rings C (X ),
we may entirely identify ﬁnitely generated projective bundles with a basic and important class of geometric objects, namely vector bundles.
There is a special case of this result which is almost immediately evident. Namely,
suppose that E is a trivial vector bundle on X , i.e., up to isomorphism E is simply
X × K n with π = π1 . Thus a section σ is nothing else than a continuous function
σ : X → K n , which in turn is nothing else than an ntuple (f1 , . . . , fn ) of elements
of C (X ). Thus if we deﬁne σi ∈ Γ(E ) simply to be the section which takes each
point to the ith standard basis vector ei of K n , we see immediately that (σ1 , . . . , σn )
is a basis for Γ(E ), i.e., Γ(E ) is a free C (X )module of rank n. Moreover, we have
Hom(X × K n , X × K m ) ∼ Map(X, HomK (K n , K m ))
=
∼ C (X ) ⊗K Hom(K n , K w ) ∼ HomC (X ) (Γ(X × K n ), Γ(X × K m )).
=
=
Thus we have established that Γ gives an additive equivalence from the category of
trivial vector bundles on X to the category of ﬁnitely generated free C (X )modules.
We wish to promote this to an equivalence from locally trivial vector bundles (i.e.,
all vector bundles) to ﬁnitely generated projective modules. Oh, if only we had
some nice “geometric” characterization of ﬁnitely generated projective modules! COMMUTATIVE ALGEBRA 103 But we do: namely Proposition 19 characterizes ﬁnitely generated projective modules over any commutative ring R as being precisely the images of projection operators on ﬁnitely generated free modules. Thus the essence of what we want to show
is that for any vector bundle E over X (a compact space), there exists a trivial
vector bundle T and a projection P : T → T – i.e., an element of Hom(T, T ) with
P 2 = P such that the image of P is a vector bundle isomorphic to E . Indeed,
if we can establish this, then just as in the proof of 19 we get an internal direct
sum decomposition T = P (T ) ⊕ (1 − P )(T ) and an isomorphism P (T ) ∼ E , and
=
applying the additive functor Γ this gives us that Γ(E ) is isomorphic to a direct
summand of the ﬁnitely generated free C (X )module Γ(T ). A little thought shows
that in fact this proves the entire result, because we have characterized Vec(X ) as
the “projection category” of the additive category trivial vector bundles, so it must
be equivalent to the “projection category” of the equivalent additive category of
ﬁnitely generated free C (X )modules. So from this point on we can forget about
projective modules and concentrate on proving this purely topological statement
about vector bundles on a compact space.26
6.3. Proof of Swan’s Theorem.
Unfortunately the category of vector bundles over X is not an abelian category.
In particular, it can happen that a morphism of vector bundles does not have either a kernel or image. Swan gives the following simple example: let X = [0, 1],
E = X × K the trivial bundle, and f : E → E be the map given by f (x, y ) = (x, xy ).
Then the image of f has rank one at every x ̸= 0 but has rank 0 at x = 0. Since X
is connected, a vector bundle over X should have constant rank function. Exactly
the same considerations show that the kernel of f is not a vector bundle. However,
nothing other than this can go wrong, in the following sense:
Proposition 154. For a morphism f : E → E ′ of vector bundles over X , TFAE:
(i) The image of f is a subbundle of E ′ .
(ii) The kernel of f is a subbundle of E .
(iii) The function x → dimK (Im f )x is locally constant.
(iv) The function x → dimK (Ker f )x is locally constant.
Proof. Step 1: We ﬁrst wish to prove a special case: namely that if f : E → E ′ is
a monomorphism of vector bundles (i.e., it induces an injection on all ﬁbers) then
(Im f ) is a subbundle of E ′ and f : E → (Im f ) is an isomorphism. The issues of
whether Im f is a vector bundle and f is an isomorphism are both local ones, so it
suﬃces to treat the case where E and E ′ are trivial bundles. Suppose E ′ = X × V ,
and let x ∈ X . Choose Wx ⊂ V a subspace complementary to (Im f )x . Then
G := X × Wx is a subbundle of E ; let ι : G → E be the inclusion map. Deﬁne
θ : E ⊕ G → E ′ by θ((a, b)) = f (a) + ι(b). Then θx is an isomorphism, so there
exists an open neighborhood U of x such that θU is an isomorphism. Since E is a
subbundle of E ⊕ G, θ(E ) = f (E ) is a subbundle of θ(E ⊕ G) = E ′ on U .
Step 2: Since the rank function on a vector bundle is locally constant, (i) =⇒
26We note that [Sw62] takes a more direct approach, for instance proving by hand that the
global section functor Γ is fully faithful. In our use of projection operators and projection categories
to prove Swan’s theorem we follow Atiyah [At89, §1.4]. Aside from being a bit shorter and slicker,
this approach really brings life to Proposition 19 and thus seems thematic in a commutative
algebra course. But it is not really more than a repackaging of Swan’s proof. 104 PETE L. CLARK (iii), (ii) =⇒ (iv), and (by simple linear algebra!) (iii) ⇐⇒ (iv).
(iv) =⇒ (i): Again the issue of whether Im f is a vector bundle is a local one,
so we may assume that E = X × V is a trivial bundle. For x ∈ X , let Wx ⊂ V
be a complementary subspace to (Ker f )x . Let G = X × Wx , so that f induces
a homomorphism ψ : G → E ′ whose ﬁber at x is a monomorphism. Thus ψ is a
monomorphism on some neighborhood U of x, so ψ (G)U is a subbundle of E ′ U .
However Ψ(G) ⊂ f (E ), and since f (E ) has constant rank, and
dim ψ (G)y = dim ψ (G)x = dim f (E )x = dim f (E )y
for all y ∈ U , ψ (G)U = f (E )U . so f (E ) is a subbundle of E ′ .
(iv) =⇒ (ii): here we exploit dual bundles. The hypothesis implies that the
kernel of f ∨ : (E ′ )∨ → E ∨ has constant rank function. Since E ∨ → Coker f ∨ is
an epimorphism, (Coker f ∨ )∨ → E ∨∨ is a monomorphism: by Step 1, its image is
a subbundle. But the natural map E → E ∨∨ is an isomorphism, the restriction of
which to Ker f gives an isomorphism to the vector bundle (Coker f ∨ )∨ . So Ker f
is a vector bundle.
The proof yields the following additional information.
Corollary 155. For any morphism of vector bundles, the rank function of the
image is upper semicontinuous: that is, for any x ∈ X , there exists a neighborhood
U of x such that for all y ∈ U , dimK (Im f )y ≥ dimK (Im f )x .
Exercise 6.1: Prove Corollary 155.
Proposition 156. Let E be a vector bundle over X , and let P ∈ End(E ) =
Hom(E, E ) be a projection, i.e., P 2 = P . Then:
a) We have Ker(P ) = Im(1 − P ).
b) Im P and Im(1 − P ) are both subbundles of E .
c) There is an internal direct sum decomposition E = Im P ⊕ Im(1 − P ).
Proof. a) For all x ∈ X linear algebra gives us an equality of ﬁbers Ker(P )x =
Im(1 − P )x . This suﬃces!
b) From part a) we deduce an equality of rank functions
rIm P + rIm(1−P ) = rE .
By Corollary 155, for all x ∈ X , there is a neighborhood U of x on which rIm P is
at least as large as rIm P (x), rIm(1−P ) is at least as large as rIm(1−P ) (x) and rE is
constantly equal to rE (x). On this neighborhood the ranks of Im P and Im(1 − P )
must be constant, and therefore by Proposition 154 Im P and Im(1 − P ) are both
subbundles.
c) Again it is enough to check this ﬁber by ﬁber, which is simple linear algebra.
An inner product on a ﬁnitedimensional Rvector space V is, as usual, a symmetric
Rbilinear form ⟨, ⟩ : V × V → R which is positive deﬁnite in the sense that for
all x ∈ V \ {0}, ⟨x, x⟩ = 0. An inner product on a ﬁnitedimensional Cvector
space V is a positive deﬁnite sesquilinear form: i.e., it is Clinear in the ﬁrst variable, conjugatelinear in the second variable and again we have ⟨x, x⟩ > 0 for all
x ∈ V \ {0}.
Now let E be a K vector bundle on X . An inner product on E is a collection of
inner products ⟨, ⟩x : Ex × Ex → K on each of the ﬁbers which vary continuously in COMMUTATIVE ALGEBRA 105 x. Formally this means the following: let E ×X E be the subset of (e1 , e2 ) ∈ E × E
such that π (e1 ) = π (e2 ); then such a ﬁberwise family of inner products deﬁnes a
function from E ×X E to K , and this function is required to be continuous.
Let us say that a metrized vector bundle E on X is a vector bundle together
with an inner product. (Again, this is an abuse of terminology: we do not speak of
the inner product by name.)
Proposition 157. Let E be a metrized line bundle on X .
a) If E ′ is a subbundle of E , ﬁberwise orthogonal projection onto E ′ deﬁnes a
projection operator P ∈ End(E ) with image E ′ .
b) All short exact sequences 0 → E ′ → E → E ′′ → 0 of vector bundles are split.
c) If M is another vector bundle on X and there exists an epimorphism of bundles
q : E → M , then M is isomorphic to the image of a projection operator on E .
Proof. a) This is mostly a matter of understanding and unwinding the deﬁnitions,
and we leave it to the reader.
b) Let P be orthogonal projection onto E ′ . The restriction of the map E → E ′′ to
Ker P is an isomorphism of vector bundles. The inverse of this isomorphism gives
a splitting of the sequence.
c) By Proposition 154, since Im q = M is a vector bundle, so is Ker q , whence a
short exact sequence
0 → Ker q → E → M → 0.
By part b), there exists a splitting σ : M → E of this sequence. Then, as usual,
∼
P = σ ◦ q is a projection operator on E and q Im P : Im P → M .
Proposition 158. If X is a paracompact topological space, then every vector bundle
over X admits an inner product.
Proof. This is a rather standard topological argument which we just sketch here.
Let M be a vector bundle on X , and let {Ui }i∈I be an open covering of X such
that the restriction of M to each Ui is a trivial bundle. On a trivial bundle there
is an obvious inner product, say ⟨, ⟩x . Now, since X is paracompact, there exists a
partition of unity {φi }i∈I subordinate to the open covering {Ux }: that is,
• each φi : X → [0, 1] is continuous,
• for all x ∈ X we have supp(φi ) ⊂ Ui
• for all x ∈ X there exists an open neighborhood V of x on which all but ﬁnitely
many φi ’s vanish identically, and
∑
• for all x ∈ X , i∈I φi (x) = 1.27
Then, for x ∈ X and e1 , e2 ∈ Mx , deﬁne
∑
⟨e1 , e2 ⟩x :=
φi (x)⟨e1 , e2 ⟩i ;
i the sum extends over all i ∈ I such that x ∈ Ui . This is an inner product on M .
To complete the proof of Swan’s Theorem, it suﬃces to show that if X is compact, every vector bundle M on X is the epimorphic image of a trivial bundle. In
particular, Proposition X.X then shows that M is a direct summand of a trivial
vector bundle T and thus Γ(M ) is a direct summand of the ﬁnitely generated free
C (X )module Γ(T ), hence is ﬁnitely generated projective.
27See e.g. Exercise 5 in §4.5 of Munkres’ Topology: a ﬁrst course for a proof of this fact. 106 PETE L. CLARK Proposition 159. Let X be a compact space and M a vector bundle on X . Then
there exists an epimorphism of bundles from a trivial vector bundle X × V to M .
Proof. Step 1: We claim that for each x ∈ X , there exists a neighborhood Ux of x
and ﬁnite set of global sections Sx = {sx,1 , . . . , sx,kx } of M such that for all y ∈ U ,
sx,1 (y ), . . . , sx,kx (y ) is a K basis for My .
proof of claim: Let U be an open neighborhood of x on which M is a trivial
bundle. Certainly then there exist ﬁnitely many sections s1 , . . . , sn of M over U
which when evaluated at any y ∈ U give a basis of My . We need to show that there
exists an open set W with x ∈ W ⊂ U and global sections s′ , . . . , s′ such that
n
1
for all i, s′ W = si W . For this it suﬃces to work one section at a time: let s be
i
a section of M over U . Since X is paracompact, it is normal, so there exist open
neighborhoods W and V of x with W ⊂ V , V ⊂ U . By Urysohn’s Lemma, there is
a continuous function ω : X → [0, 1] such that ω W ≡ 1 and ω X \V ≡ 0. If we then
deﬁne s′ : X → M by s′ (y ) = ω (y )s(y ) for y ∈ U and s′ (y ) = 0 for y ∈ X \ U , then
this s does the job.
Step 2: By compactness of X , there exists a ﬁnite subset I of X such that {Ux }x∈I
∪
covers X . So S = i∈I Sx is a ﬁnite set of global sections of M which when
evaluated at any x ∈ X , span the ﬁber Mx . So the K subspace V of Γ(M ) spanned
by S is ﬁnitedimensional. We deﬁne a map q : X × V → M by q (x, s) = s(x).
This is a surjective bundle map from a trivial vector bundle to M !
Remark: In the above proof the paracompactness of X seems to have been fully
exploited, but the need for compactness is less clear. In fact, at the end of [Sw62],
Swan remarks that if you replace the last step of the proof by an argument from
Milnor’s 1958 lecture notes Diﬀerential Topology, one gets a categorical equivalence
between vector bundles with bounded rank function on a paracompact space X and
ﬁnitely generated projective C (X )modules.
Remark: A more straightforward variant of Swan’s theorem concerns the case where
X is a compact diﬀerentiable manifold (say of class C ∞ ). In this case the equivalence is between diﬀerentiable K vector bundles on X and modules over the ring of
K valued C ∞ functions. Looking over the proof, one sees that the only part that
needs additional attention is the existence of diﬀerentiable partitions of unity. Such
things indeed exist and are constructed in many of the standard texts on geometry
and analysis on manifolds. We recommend [Wel80], which has a particularly clear
and complete discussion.
6.4. Applications of Swan’s Theorem.
6.4.1. Vector bundles and homotopy.
Vector bundles on a space are of interest not only to diﬀerential topologists and
geometers but also to algebraic geometers. This is because pullback of vector bundles behaves well under homotopy.
First, suppose that f : X → Y is a continuous map of topological spaces and
π : E → Y is a vector bundle on Y . We may pullback π to a vector bundle
πX : E ×Y X → X just by taking E ×Y X to be the ﬁber product of the maps
f and π , namely the subspace of X × E consisting of all pairs (x, v ) such that COMMUTATIVE ALGEBRA 107 f (x) = π (v ) ∈ Y . The map πX : E ×Y X → X is just restriction of the projection
map: (x, v ) → x.
Exercise 6.2: Show that πX : E ×Y X → X is indeed a vector bundle on X .
We abbreviate it by either f ∗ π or (more abusively) f ∗ E .
Exercise 6.3: Show that the pullback of any trivial bundle is a trivial bundle.
Theorem 160. (Covering Homotopy Theorem) Let X and Y be topological spaces
with X paracompact. Let π : E → Y be a vector bundle on Y , and let f, g : X → Y
be homotopic maps. Then the pullbacks f ∗ π and g ∗ π are isomorphic vector bundles
on X .
Proof. See for instance [Hus66, Thm. 4.7].
For our applications, it is enough to know that compact spaces are paracompact.
But for culture we also remark that any regular σ compact space is paracompact,
e.g. any CWcomplex with only ﬁnitely many cells of any given dimension.
Corollary 161. If X is a contractible paracompact space, then every vector bundle
on X is trivial.
Proof. Choose any point x0 ∈ X , let f : X → X be the map which sends every
point of X to x0 ,and let g : X → X be the identity map. If π : E → X is any
vector bundle on X , then by Theorem 160 we have f ∗ π = g ∗ π . Since g is the
identity map, g ∗ π = π . On the other hand, tracking through the deﬁnitions shows
f ∗ π = X × π −1 (x0 ), a trivial bundle. So π is trivial.
6.4.2. Projective Modules over C ([0, 1]).
Let [0, 1] be the closed unit interval with its topology: a compact, contractible
space. By Corollary 161, every vector bundle over [0, 1] is trivial. By Swan’s Theorem, this implies that every ﬁnitely generated projective module over the ring
R = C ([0, 1]) of continuous functions f : [0, 1] → R is free.
But now – as in §3.9 – consider the ideal I of all functions f ∈ R which vanish near zero, i.e., for which there exists ϵ(f ) > 0 such that f [0,ϵ(f )] ≡ 0. By
Exercise X.X, I is a projective Rmodule. Moreover, I is not a free Rmodule:
indeed, any f ∈ I is annihilated by any continuous function with support lying in
[0, ϵ(f )], and nonzero such functions clearly exist. On the other hand, any nonzero
free module has elements with zero annihilator: take any basis element.
Thus C ([0, 1]) is a connected ring over which every ﬁnitely generated projective
module is free, but the inﬁnitely generated projective module I is not free. (Recall
that Theorem 25 says that no such modules exist over connected Noetherian rings.)
Moreover I is therefore clearly not a direct sum of ﬁnitely generated modules, since
by what we have established any such module over C ([0, 1]) would be free!
Exercise 6.4: Use Corollary 56 to give a purely algebraic proof that I is not a
direct sum of ﬁnitely generated submodules. 108 PETE L. CLARK Exercise 6.5*: Find necessary and suﬃcient conditions on a compact, contractible
space X for there to exist a nonfree projective module.
6.4.3. Some stably free modules.
For n ∈ N, deﬁne
Rn := R[x0 , . . . , xn ]/(x2 + . . . + x2 − 1).
0
n
N
Recalling that every Rn module map from Rn → RM uniquely corresponds to a
n
matrix M ∈ MM,N (Rn ), consider the map H : Rn+1 → Rn associated to the matrix
(x0 · · · xn ). The image of H is an ideal of Rn . Applying H to each of the standard
n
basis vectors e0 , . . . , en of Rn+1 we see that x0 , . . . , xn all lie in the image of H ,
2
2
hence so does x0 + . . . + xn = 1: that is, H is surjective. We deﬁne Pn = ker(H ),
so we have a short exact sequence
H n
0 −→ Pn −→ Rn+1 −→ Rn −→ 0. Since R is free, this sequence splits: indeed, in this case we have a canonical splitting
n
given by the section σ : Rn → Rn+1 with associated matrix (x0 · · · xn )t . Thus
∼
Pn ⊕ Rn = Rn+1 ,
n so Pn is projective and moreover stably free (recall a module M is stably free if
there exist ﬁnitely generated free modules F1 and F2 with M ⊕ F1 ∼ F2 ). It is
=
natural to ask whether Pn is in fact a free Rn module. Here is the answer.
Theorem 162. (Swan) The stably free Rn module Pn is free iﬀ n = 0, 1, 3 or 7.
Let us sketch the proof of this theorem.
Step 0: To show that a module is free, it is of course enough to ﬁnd a basis.
Note that P0 = 0, so this is a trivial case. (It will become clear later why this
case was included.) When n = 1, it is an easy direct calculation to show that
−x1 e0 + x0 e1 is a basis for P1 . Similarly concrete formulas can be found for bases
of Pn when n = 3 and n = 7. We leave this as an exercise for the reader.
Step 1: Now suppose that n is not 0, 1, 3 or 7. We wish to show that Pn is not
free. For this it is suﬃcient to ﬁnd some extension ring S of R such that Pn ⊗R S is
not free. So now is a good time to observe that Rn is nothing else than the ring of
polynomial functions on the unit nsphere S n inside Euclidean space Rn+1 . That is,
every polynomial f ∈ R[x0 , . . . , xn ] deﬁnes by restriction a realvalued function on
S n , but we wish to regard two such polynomials as giving the same function if they
agree at every point of S n , but the ideal of functions vanishing identically on S n is
precisely I = (x2 + . . . + x2 − 1).28 Since every polynomial function is continuous
n
0
in the usual Euclidean topology on S n , we get an extension of rings Rn ⊂ C (S n ).
So it is enough to show that the ﬁnitely generated projective C (S n )module
Tn := Pn ⊗Rn C (S n )
is not free.
28This part of the argument will become more clear when we study aﬃne rings and the Nullstellensatz later on. Indeed the polynomial x2 + . . . + x2 − 1 is irreducible when n > 0 so the ideal
n
0
I is prime, thus radical, so I (V (I )) = I . COMMUTATIVE ALGEBRA 109 Step 2: Now we are in a Swan’s Theorem situation: the module Tn corresponds to
a unique vector bundle on S n . In fact we claim that it corresponds to the tangent
bundle T S n on S n . Since we have not given a rigorous deﬁnition or construction
of the tangent bundle to a manifold, we are not in a position to prove this here,
but here are some hints in this direction. We have S n ⊂ Rn+1 . The tangent bundle
to Rn+1 is of course trivial, hence so is its pullback – here, restriction – to S n , say
F n+1 . Further, there is a surjective bundle map from F n+1 to the trivial bundle of
rank one, say F 1 : at every point of the nsphere we take the outward unit normal
vector. The kernel of this bundle map is the tangent bundle to S n . Thus we have
a short exact sequence of vector bundles:
0 −→ T S n −→ F n+1 −→ F 1 −→ 0.
We claim that under the equivalence of Swan’s Theorem, this split exact sequence
corresponds to the split exact sequence
H 0 −→ Tn −→ C (S n )n+1 −→ C (S n ) −→ 0,
the base change to C (Sn ) of the short exact sequence of Rn modules deﬁning Pn .
Step 3: We invoke the following classical theorem.
Theorem 163. (BottMilnor [BM58]) The tangent bundle to the nsphere is trivial
iﬀ n = 0, 1, 3 or 7.
This completes the proof of Theorem 162!
Theorem 163 is of course a deep and celebrated result: the original proof used
recently developed properties of StiefelWhitney classes, Pontrjagin classes, cohomology operations, and so forth. In 1962 J.F. Adams solved a more general problem:
for each n he found the largest rank k of a trivial subbundle of the tangent bundle to S n [Ad62]. Somewhat more recently it was found that Ktheory gives more
graceful proofs: we recommend that the reader consult for instance [Ka08, §V.2].
If one merely wants some values of n for which Pn is not free, one can use much
lowerpowered diﬀerential topological machinery. Indeed, the Poincar´Hopf theoe
rem [Mi65, p. 35] shows that a closed nmanifold can only admit an everywhere
nonvanishing vector ﬁeld (this is much weaker than triviality of its tangent bundle,
which involves ﬁnding n everywhere linearly independent vector ﬁelds) if its Euler
characteristic is 0. But the Euler characteristic of S n is 1 + (−1)n , so is nonzero for
all even n. Thus BottMilnor is not needed to show that Pn is nonfree for all even n.
One can of course also ask whether all this topological stuﬀ is really necessary
to prove the nonfreeness of the stably free modules Pn . The answer is: apparently,
yes! So far as I know, there is no purely algebraic proof of Theorem 162.
7. Localization
7.1. Deﬁnition and ﬁrst properties.
As we have seen, one way to “simplify” the study of ideals in a ring R is to pass to
a quotient ring R/I : as we have seen, this has the (often useful) eﬀect of “cutting
oﬀ the bottom” of the ideal lattice by keeping only ideals J ⊃ I . There is another 110 PETE L. CLARK procedure, localization, which eﬀects the opposite kind of simpliﬁcation: given a
prime ideal P of R, there is a ring RP together with a canonical map ι : R → RP
such that ι∗ : I (RP ) → I (R) is an injection whose image is precisely the ideals
J ⊂ P . As usual, ι∗ carries prime ideals to prime ideals. In particular, assuming
only that P is prime, we get a corresponding ideal – rather inelegantly but standardly denoted P RP – which is the unique maximal ideal of RP . If we can take
P = (0) – i.e., if R is a domain – this means that P RP is the only ideal of RP ,
which is therefore a ﬁeld. In fact it is nothing else than the quotient ﬁeld of the
integral domain R, and – with one exception – all the secrets of localization are
already present in this very familiar special case.
In fact the localization construction is a bit more general than this: given an arbitrary ring R (of course commutative with unity!) and an arbitrary multiplicative
subset S of R – this just means that 1 ∈ S and SS ⊂ S – we will deﬁne a new ring
RS together with a canonical homomorphism ι : R → RS (for which ι∗ will still be
an injection with explicitly given image). In fact, just as in the case of quotients,
ι satisﬁes a certain universal mapping property, but let us sacriﬁce some elegance
for intelligibility by working our way up to this crisp deﬁnition.
Indeed, ﬁrst consider the special case in which R is a domain, with fraction ﬁeld
F . Then RS will be an extension ring of R, still with fraction ﬁeld F , which is
obtained by adjoining to R all elements 1 for s ∈ S .
s
Example: Suppose R = Z, S = {2n }n∈N . Then RS = Z[ 1 ]. Indeed we see that for
2
any nonzero element f , we can take S to be the multiplicative set consisting of the
1
powers of f , and then the localization is just R[ f ].
What if in the example above, instead of taking the multiplicative subset generated by 2, we took the multiplicative subset generated by 22 , or 2127 ? Clearly it
k −1
wouldn’t matter: if we have 21 for any k in our subring of Q, we also have 2 2k = 1 .
k
2
To generalize this idea, deﬁne the saturation S of a multiplicatively closed subset
S of a domain R to be the set {a ∈ R  ∃b ∈ R  ab ∈ S }, i.e., the set of all divisors
of elements of S . The same observation as above shows that RS = RS , so if we like
we can restrict to consideration of saturated multiplicatively closed subsets.
Example, continued: The saturated, multiplicatively closed subsets of Z correspond to (arbitary) subsets P of the prime numbers (exercise!). In particular Z
1
itself corresponds to P = ∅, Z[ p ] corresponds to P = {p}, Q corresponds to the set
of all primes. Most interestingly, ﬁx any prime p and let P be the set of all primes
except p: then the corresponding ring, which is confusingly denoted Z(p) is the
set of all rational numbers of the form x where p does not divide y . Notice that
y
such rings are the maximal subrings of Q which are not ﬁelds. Moreover, the units
of Z(p) are precisely the elements of the form x with (p, x) = 1. The nonunits a
y
are all of the form pa′ for a′ ∈ Z(p) , so therefore the unique maximal ideal is the
principal ideal (p) = pZ(p) .
Exercise 7.1: Show that the only ideals in Z(p) are those of the form (p)k for
some k ∈ N. Notice that this set happens to be identiﬁable with the set of all ideals COMMUTATIVE ALGEBRA 111 I of Z which are disjoint from the multiplicative set S (P ).
Now let R be any ring and S a multiplicatively closed subset of R. We would
still like to deﬁne a ring S −1 R which is, roughly speaking, obtained by adjoining
to R all inverses of elements of S . We can still deﬁne S −1 R in terms of formal
quotients, i.e., as equivalence classes of elements (a, b) with a ∈ R, b ∈ S . However,
if we deﬁne (a, b) ∼ (c, d) to be ad = bc, then unfortunately we ﬁnd that this need
not be an equivalence relation! Therefore we need to enlarge the relation a bit: we
put (a, b) ∼ (c, d) iﬀ there exists s ∈ S such that sad = sbc. We then deﬁne
ab
at + bs
+ :=
,
s
t
st
ab
ab
· := .
st
st
We must check that these operations are welldeﬁned on equivalence classes; this is
left as a (perhaps somewhat tedious, but not diﬃcult) exercise for the reader.
Exercise 7.2: Indeed, check that S −1 R is a ring and that x → x deﬁnes a ho1
momorphism of rings R → S −1 R. Thus S −1 R is an Ralgebra, and in particular
an Rmodule.
Exercise 7.3: Let R be a domain and S = R• = R \ {0}. Show that S −1 R is
indeed the fraction ﬁeld of R.
When f ∈ R, we denote the localization of R at the multiplicative subset generated by f as Rf .
Example: Suppose f ∈ R is a nilpotent element: f n = 0 for some n ∈ Z+ . Then
f n−1
0
1 = f n−1 whereas 0 = f . Since (f n−1 · f − f n−1 · 0) = 0, we have that 1 = 0, i.e.,
Rf is the zero ring. Conversely, if f is not nilpotent, then if it is a unit, Rf = R.
1
Otherwise, all the powers of f are distinct, and then f ̸= 1 , since for any n ∈ N,
1
n
f (1 − f ) ̸= 0, so that Rf is not the zero ring. In general, S −1 R is the zero ring
iﬀ S contains 0. This is to be regarded as a trivial case, and may safely be tacitly
excluded in the sequel.
Exercise 7.4: a) Show that the kernel of the natural map R → S −1 (R) is the
set of all r ∈ R such that for some s ∈ S , sr = 0.
b) The map R → S −1 (R) is injective iﬀ S has no zerodivisors.
c) Show that the subset Q of all nonzerodivisors of a ring R is multiplicatively
closed. The localization Q−1 R is called the total fraction ring of R. Show that
Q−1 (R) is a ﬁeld iﬀ R is an integral domain.
Exercise 7.5: Show that the homomorphism R → S −1 R is universal for homomorphisms R → T with f (S ) ⊂ T × .
7.2. Pushing and pulling via a localization map.
Let R be a ring and S a multiplicatively closed subset. Let ι : R → S −1 R be 112 PETE L. CLARK the natural map. As for any homomorphism of rings, ι induces maps between the
sets of ideals of R and the set of ideals of S −1 R, in both directions:
ι∗ : IR → IS −1 R , I → IS −1 R,
ι∗ : IS −1 R → IR , J → ι−1 (J ).
Lemma 164. Let ι : R → S −1 R be a localization map. Then for any ideal I of R,
x
ι∗ (I ) = { ∈ S −1 R  x ∈ I, s ∈ S }.
s
Proof. Let us temporarily write
x
I = { ∈ S −1 R  x ∈ I, s ∈ S }.
s
We want to show that I = ι∗ (I ) = ⟨ι(I )⟩S −1 R . It is clear that ι(I ) ⊂ I ⊂ ι∗ (I ), so
1
2
it is enough to show that I is itself an ideal of S −1 R. No problem: if x1 , x2 ∈ I ,
s
s
x1
x2
x1 s2 + x2 s1
+
=
∈ I,
s1
s2
s1 s2
and if y
s ∈ S −1 R, then
y x1
x1 y
=
∈ I.
s s1
ss1 Like quotient maps, any localization map has the pullpush property.
Proposition 165. Let ι : R → S −1 R be a localization. For any ideal J of S −1 R,
J = ι∗ ι∗ J.
Proof. We have seen before that for any homomorphism ι : R → R′ of rings and
any ideal J of R′ we have
J := ι∗ ι∗ J ⊂ J.
Thus it is enough to show the reverse containment. For this, consider an arbitrary
element x ∈ J . Then x = s x ∈ J hence also x ∈ ι∗ (J ), so ι(x) ∈ J . But since J is
s
s
an ideal and s is a unit in S −1 R, we then also have 1 x = x ∈ J .
s
s
Lemma 166. Let ι : R → S −1 R be a localization map and I an ideal of R. TFAE:
(i) I ∩ S ̸= ∅.
(ii) ι∗ (I ) = S −1 R.
Proof. (i) =⇒ (ii): If s ∈ S ∩ I , then s ∈ IS −1 R, so 1 = s ∈ ι∗ (I ).
s
(ii) =⇒ (i): Suppose 1 ∈ ι∗ (I ). By Lemma 164, 1 = x for some x ∈ I and s ∈ S .
1
s
Clearing denominators, there is s′ ∈ S such that ss′ = s′ x and thus ss′ ∈ I ∩ S .
Proposition 167. Let ι : R → S −1 R be a localization homomorphism.
a) For a prime ideal p of R, TFAE:
(i) The pushforward ι∗ p is prime in S −1 R.
(ii) The pushforward ι∗ p is proper in S −1 R.
(iii) The sets p ∩ S = ∅.
b) If p is prime and disjoint from S , then ι∗ (ι∗ p) = p. COMMUTATIVE ALGEBRA 113 Proof. a) By Lemma 166, (ii) ⇐⇒ (iii) for all ideals of R.
(i) =⇒ (iii): By contraposition. Once again, if s ∈ p ∩ S , then ι∗ (p) = S −1 R: that
is, it is not even proper, let alone prime.
1
2
(iii) =⇒ (i): Suppose p is a prime ideal of R, and suppose we have a1 , a2 ∈ S −1 R
s
s
a1 a2
x
′
with s1 s2 = s ∈ ι∗ (p). Clearing denominators, there is s ∈ S such that
ss′ a1 a2 = s′ s1 s2 x ∈ p.
Since S ∩ p = ∅, (ss′ ) ∈ p, and since p is prime, we conclude that a1 a2 ∈ p and
/
i
then that ai ∈ p for some i, hence ai ∈ ι∗ p for some i and ι∗ (p) is prime. This
s
completes the proof of part a).
b) Recall: for any homomorphism ι : R → R′ and any ideal I of R we have
ι∗ (ι∗ (I )) ⊃ I,
so taking I = p to be prime it suﬃces to show the inverse inclusion. Suppose
x ∈ ι∗ ι∗ p, i.e., there exist a ∈ p, s ∈ S such that ι(x) = x = a . By deﬁnition, this
1
s
means that there exists some s′ ∈ S such that s′ sx = s′ a ∈ p. Therefore either
s′ s ∈ p or x ∈ p, but since s′ s ∈ S and S is disjoint from p, we must have x ∈ p.
Corollary 168. The maps ι∗ and ι∗ give mutually inverse bijections from the set
of prime ideals of S −1 R to the set of prime ideals of R which are disjoint from S .
Therefore we may – and shall – view Spec S −1 R as a subset of Spec R.
Exercise 7.6:
a) Show that the results of Proposition 167 extend to all primary ideals of R.29
b) Let I be any ideal of R. Show that
ι∗ ι∗ I = {x ∈ R  ∃s ∈ S such that sx ∈ I }.
Comment: in class I remarked that there is no nice pushpull formula for an arbitrary ideal in a localization map. Whether this exercise contradicts that assertion
depends upon how nice you ﬁnd this formula to be! Try it out on the following:
c) Exhibit a map ι : R → S −1 R and a (nonprimary) ideal I of R such that ι∗ ι∗ I I .
7.3. The ﬁbers of a morphism.
Let f : R → S be a homomorphism of rings, and let p ∈ Spec R. Consider the
“ﬁber of f ∗ : Spec S → Spec R over p”, i.e.,
fp = (f ∗ )−1 (p) = {P ∈ Spec S  f ∗ (P ) = p}.
We claim that fp is canonically isomorphic to the spectrum of a certain ring.
Namely, let k (p) be the fraction ﬁeld of the domain R/p. Then we wish to identify
fp with Spec(S ⊗R k (p)).
Let ι1 : S → S ⊗R k (p) and ι2 : k (p) ⊗R S ⊗R k (p) be the canonical maps. The tensor product of Ralgebras ﬁts into a commutative square (INSERT) and is indeed
the categorical pushout: in other words, given any ring A and homomorphisms
φ1 : A → S φ2 : A → k (p) such that the composite homomorphisms ι1 ◦ φ1 = ι2 ◦ φ2
are equal, there exists a unique homomorphism Φ : A → R such that f ◦ Φ = φ1
29Recall p is primary if for a, b ∈ R such that ab ∈ p, either a ∈ p or bn ∈ p for some n ∈ Z+ .
We have not yet done much with this concept, and will not really address it squarely until the
section on primary decomposition. 114 PETE L. CLARK and q ◦ Φ = φ2 .
On the spectral side, all the arrows reverse, and the corresponding diagram is
(INSERT), which expresses Spec(S ⊗R k (p)) as the ﬁber product of Spec S and
Spec k (p) over Spec R.
Observe that the map ι1 : S → S ⊗R k (p) is the composite of the surjective map
q1 : S → S ⊗R R/p with the map ℓ2 : S ⊗R R/p → (S ⊗R R/p) ⊗R/p k (p), the latter
map being localization with respect to the multiplicatively closed subset q1 (R \ p).
∗
∗
Both q1 and ℓ∗ are injections, and therefore ι1 ∗ = q1 ◦ ℓ∗ is injective. Similarly
2
2
Spec k (p) → Spec R (this is just the special case of the above with R = S ). It
follows that the above diagram identiﬁes Spec S ⊗R k (p) with the prime ideals P
of Spec S such that f ∗ P = p.
7.4. Commutativity of localization and passage to a quotient.
Lemma 169. Let R be a ring, S ⊂ R a multiplicatively closed subset, and I and
ideal of A. Write q : R → R/I for the quotient map and put S := q (S ). Then there
is a canonical isomorphism
S −1 R/IS −1 R ∼ S
= −1 (R/I ). Proof. Explicitly, we send a (mod I )S −1 R to a , where a = a + I , s = s + I . It is
s
s
straightforward to check that this an isomorphism.
Remark: Matsumura makes the following nice comment: both sides satisfy the
universal property for homomorphisms f : R → R′ such that f (S ) ⊂ (R′ )× and
f (I ) = 0. Therefore they must be canonically isomorphic.
7.5. Localization at a prime ideal.
An extremely important example of a multiplicative subset of R is the complement
R \ p of a prime ideal p. As a matter of notation, we write Rp for (R \ p)−1 R.30
Proposition 170. If p is a prime ideal of R, then the localization Rp is a local
ring with unique maximal ideal pRp .
Proof. We know that the primes of the localized ring are precisely the pushforwards
of the prime ideals of R which are disjoint from the muliplicatively closed set. Here
S = R \ p, so being disjoint from S is equivalent to being contained in p. Thus the
unique maximal such element is indeed pRp .
Remark: We will simply write p for the maximal ideal pRp of Rp .
Proposition 170, simple though it is, is of inestimable importance. It shows that
the eﬀect of localization at a prime ideal on the lattice of ideals is dual to that of
passage to the quotient: if we mod out by a prime p, we get a ring R/p whose ideals
are precisely the ideals of R containing p. However, if we localize at R \ p, we get
a ring whose ideals are precisely the ideals of R contained in p. In particular, this
30This is inevitably a bit confusing at ﬁrst, but our choice of notation for a loacalization is
designed to make this less confusing. The other common notation for the localization, RS , creates
a notational nightmare. As a mnemonic, remember that we gain nothing by localizing at a subset
S containing 0, since the corresponding localization is the trivial ring. COMMUTATIVE ALGEBRA 115 construction motivates us to develop an especially detailed theory of local rings, by
assuring us that such a theory could be put to good use in the general case.
7.6. Localization of modules. If S is any multiplicative subset and M is any Rmodule, we can also construct a localized Rmodule S −1 M . One the one hand, we
can construct this exactly as we did S −1 R, by considering the appropriate equivalence relation on pairs (m, s) ∈ M × S . On the other hand, we can just take the
base extension S −1 R ⊗ M . We are left with the task of showing that these two
constructions are “the same”.
Exercise 7.7: Formulate a universal mapping property for the localization morphism M → S −1 M . Check that both of the above constructions satisfy this universal mapping property, and deduce that they are canonically isomorphic.
Exercise 7.8: Show that the kernel of M → S −1 M is the set of m ∈ M such
that ann(m) ∩ S ̸= ∅.
Exercise 7.9: Let N be any S −1 Rmodule. Show that there exists an Rmodule M
such that N ∼ S −1 R ⊗R M .
=
Generally speaking, thinking of S −1 M as S −1 R ⊗R M is more convenient for proving results, because it allows us to employ the theory of tensor products of modules
that we developed in §X.X above. For example:
Proposition 171. For any ring R and multiplicatively closed subset S of R, S −1 R
is a ﬂat Rmodule. Equivalently, if
0 → M ′ → M → M ′′ → 0
is a short exact sequence of Rmodules, then
0 → S −1 M ′ → S −1 M → S −1 M ′′ → 0
is a short exact sequence of Rmodules (or equivalently, of S −1 Rmodules).
Proof. Tensor products are always right exact, so we need only show S −1 M ′ →
′
S −1 M . Suppose not: then there exists m′ ∈ M ′ and s ∈ S such that m = 0 ∈ M .
s
′
Thus there is g ∈ S such that gm′ = 0, but if so, then m = 0 in M ′ .31
s
Corollary 172. Let N and P be submodules of an Rmodule M . Then:
a) S −1 (N + P ) = S −1 N + S −1 P .
b) S −1 (N ∩ P ) = S −1 N ∩ S −1 P .
c) S −1 (M/N ) ∼S −1 R S −1 M/S −1 N .
=
Exercise 7.10: Prove Corollary 172.
Proposition 173. Let M and N be Rmodules and S a multiplicatively closed
subset of R. Then the mapping
m⊗n
mn
⊗→
s
t
st
31Note also that the exactness of a sequence of Rmodules does not depend on the Rmodule
structure but only on the underlying abelian group structure. Thus if we have a sequence of abelian
groups which can be viewed as a sequence of Rmodules and also as a sequence of R′ modules,
then exactness as Rmodules is equivalent to exactness as R′ modules. 116 PETE L. CLARK induces an isomorphism of S −1 (R)modules
∼ S −1 M ⊗S 1 R S −1 N → S −1 (M ⊗R N ).
In particular, for any prime ideal p of R, we have
∼ Mp ⊗Rp Np → (M ⊗R N )p .
Exercise 7.11: Prove Proposition 173.
Exercise 7.12: Let R be a ring, S a multiplicative subset of R, and M an Rmodule.
a) If M is ﬁnitely generated, then S −1 M is a ﬁnitely generated S −1 Rmodule.
b) If M is ﬁnitely presented, then S −1 M is a ﬁnitely presented S −1 Rmodule.32
7.7. Local properties.
We say that a property P of a ring R is localizable if whenever R satisﬁes property P , so does Rp for every prime ideal p of R. We say that a property P is
localtoglobal if whenever Rp has property P for all prime ideals p of R, then R
has that property. Finally, we say a property is local if it is both localizable and
localtoglobal. There are similar deﬁnitions for properties of Rmodules.
One of the most important themes in commutative algebra is the recognition of
the importance of local properties for rings and modules.
Remark: Very often it is true that if P is a local property, then R has property P iﬀ Rm has property P for all maximal ideals m of R. We will not introduce
terminology for this, but watch for it in the upcoming results.
First of all, for an Rmodule, being trivial is a local property.
Proposition 174. For an Rmodule M , TFAE:
(i) M = 0.
(ii) Mp = 0 for all primes p of R.
(iii) Mm = 0 for all maximal ideals m of R.
Proof. Clearly (i) =⇒ (ii) =⇒ (iii), so asume that Mm = 0 for all maximal ideals
m of R. Suppose there exists 0 ̸= x ∈ M , and let I be the annihilator of x, so that
I is a proper ideal of R and thus contained in some maximal ideal m. Then x is
not killed by any element of the multiplicative subset R \ m and therefore maps to
a nonzero element of Mm : contradiction.
Proposition 175. Let f : M → N be an Rmodule homomorphism.
a) TFAE:
(i) f is injective.
(ii) For all prime ideals p of R, fp : Mp → Np is injective.
(iii) For all maximal ideals m of R, fm : Mm → Nm is surjective.
b) Part a) holds with “injective” replaced everywhere by “surjective”, and thus also
if “injective” is replaced everywhere by “is an isomorphism.”
32Actually both parts hold for any base change R → R′ ! We record it in this form since it will
be used later. COMMUTATIVE ALGEBRA 117 Proof. a) (i) =⇒ (ii) by the exactness of localization, and obviously (ii) =⇒ (iii).
Assume (iii), and let M ′ = Ker(f ). Then 0 → M ′ → M → N is exact, hence for
′
′
all m we have 0 → Mm → Mm → Nm is exact. So, by our assumption, Mm = 0 for
′
all maximal ideals m, and thus by Proposition 174 we have M = 0. The proof of
part b) is virtually identical and left to the reader.
Warning: Note that Proposition 175 does not say the following: if M and N
are Rmodules such that Mp ∼ Np as Rp modules for all p ∈ Spec R, then M ∼ N .
=
=
This is being asserted only when there is a map f : M → N inducing all the isomorphisms between localized modules.
Exercise 7.13: Exhibit ﬁnitely generated Rmodules M and N which are “locally
isomorphic” – i.e., Mp ∼ Np for all p ∈ Spec R – but are not isomorphic.33
=
Corollary 176. Let R be an integral domain with fraction ﬁeld K . Then as m
∩
ranges over all maximal ideals of R, m Rm = R.
∩
Proof. Consider the injection f : R → S := m Rm . Let p and q be distinct
maximal ideals. Then Rq ⊗R Rp = K , so for any maximal ideal m, Sm = Rm
and the localized map fm : Rm → Rm is an isomorphism. Therefore f itself is an
isomorphism, i.e., surjective.
7.7.1. Local nature of ﬂatness.
Proposition 177. For an Rmodule M , TFAE:
(i) M is ﬂat.
(ii) For all prime ideals p of R, Mp is ﬂat.
(iii) For all maximal ideals m of R, Mm is ﬂat.
Proof. (i) =⇒ (ii) is a special case of Proposition 171; (ii) =⇒ (iii) is immediate.
So assume (iii), and let N → P be any injective Rmodule homomorphism. Then,
by exactness of localization, for all maximal ideals m we have Nm → Pm . Since
Mm is assumed to be ﬂat, we have (N ⊗R M )m = Nm ⊗Rm Mm → Pm ⊗Rm Mm =
(P ⊗R M )m . Applying Proposition 175 we conclude that N ⊗R M → P ⊗R P is
injective, and therefore M is ﬂat over R.
Corollary 178. Let R be a ring, S ⊂ R a multiplicative subset. If M is a ﬂat
Rmodule, then S −1 M is a ﬂat S −1 Rmodule.
Proof. If M is ﬂat, so is Mp for each prime ideal p of M , but since the primes of
S −1 R are a subset of the primes of R, this implies that S −1 M is ﬂat.
When a property P of rings or modules is not local, it is often of interest to study
also its “localized version”: we say that an Rmodule M is locally P if for all
prime ideals p of R, Mp has property p (and similarly for rings).
7.7.2. Absolute ﬂatness revisited.
Lemma 179. A local ring (R, m) is absolutely ﬂat iﬀ it is a ﬁeld.
33In mantra form: “being isomorphic” is not a local property, but “being an isomorphism” is! 118 PETE L. CLARK Proof. Certainly a ﬁeld is absolutely ﬂat. The idea of the converse is that by Proposition XX, an absolutely ﬂat ring must have many idempotent elements, whereas a
local ring can have no nontrivial idempotents. More precisely, let R be a local ring
which is not a ﬁeld, and let x be a nonzero, nonunit element. Then I = (x) is a
proper ideal, and by Proposition X.X we would have R ∼ I ⊕ J .
=
Proposition 180. Let R be a ring.
a) If R is absolutely ﬂat and S ⊂ R is any multiplicative subset, then S −1 R is
absolutely ﬂat.
b) R is absolutely ﬂat iﬀ for every maximal ideal m of R, Rm is a ﬁeld.
Proof. a) By Exercise X.X, every S −1 Rmodule is of the form S −1 R ⊗R M for some
Rmodule M . By hypothesis M is ﬂat, so by Corollary 178, so is S −1 M .
b) If R is absolutely ﬂat, and m is a maximal ideal of R, then by part a) Rm is
absolutely ﬂat. On the other hand it is a local ring, so by Lemma 179 Rm is a ﬁeld.
Conversely, assume that each Rm is a ﬁeld, and let M be an Rmodule. Then for
all m ∈ MaxSpec R, Mm is a ﬂat Rm module, so M is a ﬂat Rmodule.
Of course it is natural to ask about whether freeness and projectivity are local
properties. We devote the following section to an analysis of this question.
7.8. Local characterization of ﬁnitely generated projective modules.
7.8.1. Zlocal properties.
Let us call a family of {fi }i∈I of elements of R a Zfamily if ⟨fi ⟩ = 1. Clearly for
every Zfamily there is a ﬁnite subset J ⊂ I such that {fi }i∈J is also a Zfamily.
(Later on, this trivial observation will be dressed up in rather fancy attire: this
gives the quasicompactness of the Zariski topology on Spec R.)
A property P of rings or modules will be said to be Zlocal if it holds over R
iﬀ it holds over all Rfi for some Zfamily {fi } of R.
Proposition 181.
Let u : M → N be a homomorphism of Rmodules, and let p ∈ Spec R.
a) If N is ﬁnitely generated and up is surjective, there exists f ∈ R \ p such that
uf : Mf → Nf is surjective.
b) The surjectivity of u is a Zlocal property.
c) If M is ﬁnitely generated, N is ﬁnitely presented and up is an isomorphism, then
there exists f ∈ R \ p such that uf : Mf → Nf is an isomorphism.
d) If M is ﬁnitely generated and N is ﬁnitely presented, then the bijectivity of u is
a Zlocal property.
Proof. Write out the exact sequence
u 0 → ker u → M → N → coker u → 0.
By the ﬂatness of localization, this sequence remains exact upon being tensored
with Rf for any f ∈ R or with Rp for any p ∈ R. It follows that passage to the
kernel and cokernel commutes with localization.
a) We’re assuming 0 = coker(up ) = (coker u)p , i.e., for each x ∈ coker u there exists
fx ∈ R \ p such that fx x = 0. Since coker u is a quotient of the ﬁnitely generated
module N , it is ﬁnitely generated, say by x1 , . . . , xn . Then f = fx1 · · · fxn ∈ R \ p COMMUTATIVE ALGEBRA 119 is such that f coker u = 0, so 0 = (coker u)f = coker(uf ) and uf is surjective.
b) It is clear that if u is surjective, then for any f ∈ R, uf is surjective. Conversely,
let {fi }i∈I be a Zfamily such that ufi is surjective for all i. Then for any p ∈ Spec R
there exists i ∈ I such that fi ∈ R \ p, so that up is a further localization of ufi and
thus the surjectivity of ufi implies that of up . By Proposition 175, u is surjective.
c) By part a), there exists f1 ∈ R \ p such that coker uf1 = 0, and thus we have an
exact sequence
0 → (ker u)f1 → Mf1 → Nf1 → 0.
Since N is ﬁnitely presented over R, Nf1 is ﬁnitely presented over Rf1 and thus
(ker u)f1 is ﬁnitely generated. Arguing as in part b), we get f2 ∈ R \ p such that
∼
f1 f2 ker u = 0. Taking f = f1 f2 we get uf : Mf → Nf .
d) This is proved analogously to part b) and is left to the reader.
Corollary 182. For a ﬁnitely presented Rmodule M , TFAE:
(i) There is a Zfamily {fi }i∈I of R such that for all i ∈ I , Mfi is a free Rfi module.
(ii) For every prime ideal p of R, Mp is a free Rp module.
(iii) For every maximal ideal m of R, Mm is a free Rm module.
Proof. (i) =⇒ (ii): For each prime ideal p there exists at least one i such that
fi ∈ p; equivalently, the multiplicative subset generated by fi is contained in R \ p.
/
Thus Mp = Mfi ⊗Rfi Rp and since Mfi is free, so is Mp .
(ii) =⇒ (i): It is enough to ﬁnd for each prime ideal p an element fp ∈ R \ p such
that Mfp is free: for if so, then {fp }p∈Spec R is a Zfamily. Choose x1 , . . . , xn ∈ M
whose images in Mp give an Rp basis, and deﬁne u : Rn → M via ei → xi . Then
up is an isomorphism, so by Proposition 181c) we may choose fp ∈ R \ p such that
ufp is an isomorphism and thus Mfp is free.
(ii) ⇐⇒ (iii): this follows from Proposition 175.
Exercise 7.14: Let R1 , . . . , Rn be rings and for 1 ≤ i ≤ n, Mi a ﬁnitely generated
∏n
projective Ri module. Show that M = i=1 Mi is a ﬁnitely generated projective
∏n
R = i=1 Ri module.
We can now prove one of the major results of this text.
Theorem 183. Let R be a ring and M an Rmodule. The following are equivalent:
(i) M is ﬁnitely generated and projective.
(ii) M is ﬁnitely presented and for all m ∈ MaxSpec R, Mm is a free Rm module.
(iii) For every maximal ideal m of R, there exists f ∈ R \ m such that Mf is a
locally free Rf module of ﬁnite rank.
(iv) There exists a ﬁnite Zfamily {f1 , . . . , fn } of R such that ⟨f1 , . . . , fn ⟩ = R and
for all i, Mfi is a ﬁnitely generated free Rfi module.
Proof. (i) =⇒ (ii): Let M be ﬁnitely generated and projective. There exists a
ﬁnitely generated free module F and a surjection q : F → M . Since M is projective,
q splits and Ker(q ) is not just a submodule of F but also a quotient and thus ﬁnitely
generated. So M is ﬁnitely presented. Since projectivity is preserved by base change
and any ﬁnitely generated projective module over a local ring is free (Theorem 24),
for all maximal ideals m of R, Mm is free.
(ii) =⇒ (iii): this follows immediately from Corollary 182.
(iii) =⇒ (iv): For each m ∈ MaxSpec R, choose fm ∈ R \ m such that Mfm is a
ﬁnitely generated free Rfm module. Then {fm }m∈MaxSpec R is a Zfamily of R, and 120 PETE L. CLARK as remarked above, every Zfamily contains a ﬁnite subfamily.
∏n
(iv) =⇒ (i): Put S = i=1 Rfi and let f : R → S be the natural map.
Step 1: First note that
ker f = n
∩
i=1 ker(R → Rfi ) = n
∩ ann(fi ) = ann⟨f1 , . . . , fn ⟩ = ann R = 0, i=1 so f is injective, and thus S is an extension ring of R.
Step 2: We claim f : R → S is a faithfully ﬂat extension. Since localizations
are ﬂat and direct sums of ﬂat algebras are ﬂat, S/R is a ﬂat extension. So by
Theorem 106, it is enough to show that f ∗ : Spec S → Spec R is surjective. But
⨿n
Spec S = i=1 Spec Rfi and f ∗ (Spec Rfi ) is the subset of p ∈ Spec R such that
fi ∈ p. Since {f1 , . . . , fn } forms a Zfamily, no proper ideal can contain all the fi ’s,
/
and therefore p lies in at least one f ∗ (Spec Rfi ).
Step 3: We have ∏faithfully ﬂat ring extension f : R → S and an Rmodule M such
a
∏n
n
that M ⊗R S = i=1 Mfi is ﬁnitely generated and projective as an S = i=1 Rfi module (Exercise X.X). By Theorem 109, M is ﬁnitely generated and projective!
Corollary 184. Every ﬁnitely presented ﬂat Rmodule is projective.
Proof. Let M be a ﬁnitely presented, ﬂat Rmodule. For each maximal ideal m of
R, Mm is a ﬁnitely presented ﬂat module over the local ring Rm , hence is free by
Theorem 58. Therefore by criterion (iii) of Theorem 183, M is projective.
Corollary 185. Let M be ﬁnitely generated over the Noetherian ring R. TFAE:
(i) M is projective.
(ii) M is locally free.
(iii) M is ﬂat.
Exercise 7.15: Prove Corollary 185. Corollary 185 is the last word on ﬁnitely
generated projective modules over Noetherian rings. In the nonNoetherian case,
Corollary 184 leaves a little room for improvement: could it be true that every
ﬁnitely generated ﬂat module is projective? This is not true in general but it is
true for some important classes of nonNoetherian rings, e.g. any connected ring.
To see this we need to make a topological study of the rank function on a ﬁnitely
generated projective module. This is taken up later on in §X.X .
Theorem 186. For an Rmodule M , TFAE:
(i) M is ﬁnitely generated projective.
(ii) For all Rmodules B , the natural map
Φ : A∨ ⊗R B → HomR (A, B )
induced by (f, b) → (a → f (a)b) is an isomorphism.
(iii) The map Φ : A∨ ⊗R A → HomR (A, A) is an isomorphism.
Proof. (i) =⇒ (ii): It is enough to show that for all p ∈ Spec R, Φp is an isomorphism. Since M is ﬁnitely generated projective, it is ﬁnitely presented; moreover Rp is a ﬂat Rmodule, so by Theorem 101 we have a canonical isomorphism
HomR (M, N ) ⊗R Rp = HomRp (Mp , Np ). Also tensor products commute with base
change, so it is enough to show
Φp : A∨ ⊗Rp Bp → HomRp (Ap , Bp )
p COMMUTATIVE ALGEBRA 121 is an isomorphism. Since A is ﬁnitely generated projective, Ap is ﬁnitely generated
and free. We are thus essentially reduced to a familiar fact from linear algebra,
∼
namely the canonical isomorphism V ∨ ⊗ W → Hom(V, W ) for vector spaces over a
ﬁeld, with V ﬁnitedimensional. We leave the details to the reader as an exercise.
(ii) =⇒ (iii): This is immediate.
∑m
−1
(iii)
∑m =⇒ (i): Let Φ (1M ) = i=1 fi ⊗ ai . Then we have that for all a ∈ M , a =
i=1 fi (a)ai . By the Dual Basis Lemma, M is ﬁnitely generated projective.
7.8.2. Inﬁnitely generated locally free modules.
Let M be an Rmodule which is not necessarily ﬁnitely generated. Since projectivity is preserved by base change and by Theorem X.X every projective module
over a local ring is free, it follows that if M is projective it is locally free. What
about the converse?
It need not hold: for inﬁnitely generated modules, being locally free can be a
much weaker property. Consider:
Proposition 187. For a ring R with MaxSpec R = Spec R, TFAE:
(i) R is absolutely ﬂat.
(ii) Every Rmodule is locally free.
Proof. (i) =⇒ (ii): By Propoition 180, for m ∈ MaxSpec R, Rm is a ﬁeld, so every
Rm module is free. Since MaxSpec R = Spec R, every Rmodule is locally free.
(ii) =⇒ (i): Again by Proposition 180, if R is not absolutely ﬂat, then there
exists some m ∈ MaxSpec R such that Rm is not a ﬁeld, and thus there exists
a nonfree Rm module Mm . By Exercise X.X, there is an Rmodule M such that
M ⊗R Rm ∼ Rm and thus M is not locally free.
=
Remark: Later we will see that a ring R is absolutely ﬂat iﬀ every prime ideal of
R/ nil(R) is maximal. In particular, for every absolutely ﬂat ring we necessarily
have MaxSpec R = Spec R, so Proposition 187 is actually telling us that a ring is
absolutely ﬂat iﬀ it is “absolutely locally free”.
As we have seen, there are plenty of rings which are absolutely ﬂat but not absolutely projective. For instance an absolutely projective ring is Noetherian, so an
inﬁnite product of ﬁelds or an inﬁnite Boolean ring will carry locally free, nonprojective modules.
8. Noetherian rings
We have already encountered the notion of a Noetherian ring, i.e., a ring in which
each ideal is ﬁnitely generated; or equivalently, a ring which satisﬁes the ascending
chain condition (ACC) on ideals. Our results so far have given little clue as to the
importance of this notion. But in fact, as Emmy Noether showed, consideration of
rings satisfying (ACC) is a major unifying force in commutative algebra.
In this section we begin to see why this is the case. After giving an introductory examination of chain conditions on rings and modules, we are able to make
the key deﬁnitions of height of a prime ideal and dimension of a ring, which we
will slowly but surely work towards understanding throughout the rest of these 122 PETE L. CLARK notes. Indeed we begin by giving a reasonably complete analysis of the structure
theory of Artinian rings, which, as we will show, really is our ﬁrst order of business in attempting the systematic study of Noetherian rings, since according to the
AkizukiHopkins theorem the Artinian rings are precisely the Noetherian rings of
dimension zero. We are then able to state and prove three of the most important
and useful theorems in the entire subject. Whereas the ﬁrst theorem, the Hilbert
basis theorem, gives us a large supply of Noetherian rings, the latter two theorems,
Krull’s intersection theorem and Krull’s principal ideal theorem, are basic results
about the structure theory of Noetherian rings.
8.1. Chain conditions on posets.
Let S be a partially ordered set. S satisﬁes the Ascending Chain Condition
(ACC) if there is no inﬁnite sequence {xi }∞ of elements of S such that xi < xi+1
i=1
for all i ∈ Z+ . One says there is no inﬁnite properly ascending chain. Similarly, S
satisﬁes the Descending Chain Condition (DCC) if there is no inﬁnite sequence
{yj }∞ of elements of S such that yj > yj +1 for all j ∈ Z+ .
j =1
Of course, at the level of abstract partially ordered sets these two notions are
essentially equivalent: a partially ordered set S satisﬁes (ACC) iﬀ S op – the same
underlying set endowed with the opposite ordering x ≤′ y ⇐⇒ y ≤ x – satisﬁes
(DCC). Nevertheless in our applications the two conditions remain quite distinct.
Examples: If S is ﬁnite it satisﬁes both ACC and DCC. With the usual orderings, the positive integers Z+ satisfy DCC but not ACC, the negative integers Z−
(or equivalently, Z+ with the opposite ordering) satisfy ACC but not DCC, and
the integers Z satisfy neither.
Exercise 8.1: Let S be a poset.
a) Show that S satisﬁes (ACC) (resp. (DCC)) iﬀ there is no order embedding
Z+ → S (resp. Z− → S ).
b) Suppose S is totally ordered. Show that S satisﬁes (DCC) iﬀ it is wellordered:
i.e., every nonempty subset has a minimal element.
Note that strictly speaking it would be more accurate to speak of ascending and descending sequences, since generally speaking by a “chain” in a partially ordered set
one means any totally ordered subset. Thus both inﬁnite ascending and descending sequences are examples of chains, and the terms “ascending” and “descending”
cannot be meaningfully a applied to an arbitrary chain.
8.2. Chain conditions on modules.
Let R be a ring, and M a (left) Rmodule. It makes sense to speak of the (ACC)
and (DCC) for Rsubmodules of M . Indeed, we will call M a Noetherian module
if it satisﬁes (ACC) and an Artinian module if it satisﬁes (DCC).
Exercise 8.2: Show that an Rmodule M is Noetherian iﬀ every Rsubmodule M ′
of M is ﬁnitely generated.
Example: As a Zmodule, the integers Z are Noetherian but not Artinian. COMMUTATIVE ALGEBRA 123 Example: As a Zmodule, the group of all ppower roots of unity in the complex numbers – in other words, limn→∞ µpn – is Artinian but not Noetherian.
Notice that every ring R is naturally an Rmodule, and the Rsubmodules of R
are precisely the ideals. Thus it makes sense to say whether R is a Noetherian
or Artinian Rmodule, and – thank goodness – this is visibly consistent with the
previous terminology.
Exercise 8.3: Let M ′ ⊂ M be Rmodules, and φ : M → M/M ′ be the quotient
map. If N1 and N2 are submodules of M such that N1 ⊂ N2 , N1 ∩ M ′ = N2 ∩ M ′
and φ(N1 ) = φ(N2 ), show that N1 = N2 .
Theorem 188. Let 0 → M ′ → M → M ′′ → 0 be a short exact sequence of
Rmodules. Then M is Noetherian (resp. Artinian) iﬀ both M ′ and M ′′ are Noetherian (resp. Artinian).
Proof. We do the Noetherian case, leaving the similar Artinian case as an exercise
for the reader. First, since an inﬁnite ascending chain in a submodule or quotient
module of M gives rise to an inﬁnite ascending chain in M , if M is Noetherian,
both M ′ and M ′′ are. Conversely, suppose N1 N2 . . . is an inﬁnite ascending
chain of submodules of M . Consider the chain (Ni + M ′ )/M ′ in M ′′ = M/M ′ .
By hypothesis, this chain eventually stabilizes, i.e., for suﬃciently large i and j ,
Ni + M ′ = Nj + M ′ . Similarly, by intersecting with M ′ we get that for suﬃciently
large i and j Ni ∩ M ′ = Nj ∩ M ′ . Applying Exercise X.X we conclude Ni = Nj for
all suﬃciently large i, j .
A ring R is Noetherian if R is a Noetherian Rmodule. A ring R is Artinian if
R is an Artinian Rmodule.
Exercise 8.3.5: Let R be a ring.
a) Show that R is Noetherian iﬀ every ﬁnitely generated Rmodule is Noetherian.
b) Show that R is Artinian iﬀ every ﬁnitely generated Rmodule is Artinian.
c) Exhibit a ring R which is Noetherian but not Artinian.
d) Can you ﬁnd a ring R which is Artinian but not Noetherian?34
8.3. Semisimple modules and rings.
In this section we allow not necessarily commutative rings R. By a “module over
R” we mean a left Rmodule unless otherwise indicated.
A module M is simple if it is nonzero and has no proper, nonzero submodules.
This deﬁnition is of course made in analogy to that of a simple group, namely
a nontrivial group possessing no nontrivial proper normal subgroups. And indeed
many of the results in this and subsequent sections were ﬁrst proved in the context
of groups. It is even possible to work in a single context that simultaneously generalizes the case of groups and modules (over a not necessarily commutative ring),
34More on this later! 124 PETE L. CLARK the key concept being that of groups with operators. For more on this perspective we invite the reader to consult any suﬃciently thick allpurpose graduate level
algebra text, the gold standard here being [J1], [J2].
Exercise 8.4 (Schur’s Lemma): Let M be a simple Rmodule. Show that EndR (M )
is a division ring.
Theorem 189. For an Rmodule M , TFAE:
(i) M is a direct sum of simple submodules.
(ii) Every submodule of M is a direct summand.
(iii) M is a sum of simple submodules.
A modules satisfying these equivalent conditions is called semisimple.
⊕
Proof. (i) =⇒ (ii): Suppose M = i∈I Si , with each Si a simple submodule. For
⊕
each J ⊂ I , put MJ = i∈J Si . Now let N be an Rsubmodule of M . An easy
Zorn’s Lemma argument gives us a maximal subset J ⊂ I such that N ∩ MJ = 0.
For i ∈ J we have (MJ ⊕ Si ) ∩ N ̸= 0, so choose 0 ̸= x = y + z , x ∈ N , y ∈ MJ ,
/
z ∈ Si . Then z = x − y ∈ (Mj + N ) ∩ Si , and if z = 0, then x = y ∈ N ∩ Mj = 0,
contradiction. So (MJ ⊕ N ) ∩ Si ̸= 0. Since Si is simple, this forces Si ⊂ MJ ⊕ N .
It follows that M = MJ ⊕ N .
(ii) =⇒ (i): First observe that the hypothesis on M necessarily passes to all
submodules of M . Next we claim that every nonzero submodule C ⊂ M contains
a simple module.
proof of claim: Choose 0 ̸= c ∈ C , and let D be a submodule of C which
is maximal with respect to not containing c. By the observation of the previous
paragraph, we may write C = D ⊕ E . Then E is simple. Indeed, suppose not and
let 0 F
E . Then E = F ⊕ G so C = D ⊕ F ⊕ G. If both D ⊕ F and D ⊕ G
contained c, then c ∈ (D ⊕ F ) ∩ (D ⊕ G) = D, contradiction. So either D ⊕ F
or D ⊕ G is a strictly larger submodule of C than D which does not contain c,
contradiction. So E is simple, establishing our claim.
Now let N ⊂ M be maximal with respect to being a direct sum of simple
submodules, and write M = N ⊕ C . If C ̸= 0, then by the claim C contains a
nonzero simple submodule, contradicting the maximality of N . Thus C = 0 and
M is a direct sum of simple submodules.
(i) =⇒ (iii) is immediate.
(iii) =⇒ (i): as above, by Zorn’s Lemma there exists a submodule N of M which is
maximal with respect to being a direct sum of simple submodules. We must show
N = M . If not, since M is assumed to be generated by its simple submodules,
there exists a simple submodule S ⊂ M which is not contained in N . But since S
is simple, it follows that S ∩ N = 0 and thus N ⊕ S is a strictly larger direct sum
of simple submodules: contradiction.
Corollary 190. An Rmodule M has a unique maximal semisimple submodule,
called the socle of M and written Soc M . Thus M is semisimple iﬀ M = Soc M .
Exercise 8.5: Prove Corollary 190.
Exercise 8.6: Let N ∈ Z+ . Compute the socle of the Zmodule Z/N Z. Show
in particular that Z/N Z is semisimple iﬀ N is squarefree. COMMUTATIVE ALGEBRA 125 A not necessarily commutative ring R is left semisimple if R is semisimple as
a left Rmodule.
Theorem 191. For a nonzero not necessarily commutative ring R, TFAE:
(i) R is left semisimple.
(ii) Every left ideal of R is a direct summand.
(iii) Every left ideal of R is an injective module.
(iv) All left Rmodules are semisimple.
(v) All short exact sequences of left Rmodules split.
(vi) All left Rmodules are projective.
(vii) All left Rmodules are injective.
Proof. We will show (i) ⇐⇒ (ii), (iv) ⇐⇒ (v) ⇐⇒ (vi) ⇐⇒ (vii) and (ii)
=⇒ (vii) =⇒ (iii) =⇒ (ii), which suﬃces.
(i) =⇒ (ii) follows immediately from Theorem 189.
(iv) ⇐⇒ (v) follows immediately from Theorem 189.
(v) ⇐⇒ (vi) and (v) ⇐⇒ (vii) are immediate from the deﬁnitions of projective
and injective modules.
(ii) =⇒ (vii): Let I be a left ideal of R and f : I → M an Rmodule map. By
π1
hypothesis, there exists J such that I ⊕ J = R, so f extends to F : R = I ⊕ J →
I → M . By Baer’s Criterion, M is injective.
(vii) =⇒ (iii) is immediate.
(iii) =⇒ (ii) is immediate from the deﬁnition of injective modules.
Lemma 192. Let R be a ring and {Mj }j ∈J be an indexed family of nonzero Rmodules. The following are equivalent:
(i) I is ﬁnite and each Mj is ﬁnitely generated.
⊕
(ii) M = j ∈J Mj is ﬁnitely generated.
Proof. (i) =⇒ (ii) is left to the reader as an easy exercise.
(ii) =⇒ (i): Each Mj is isomorphic to a quotient of M , so if M is ﬁnitely generated,
so is Mj . Now let X = {x1 , . . . , xn } be a ﬁnite generating set for M , and for each
∑
1 ≤ 1 ≤ n, let xij be the j component of xi , so xi = j ∈J xij . This sum is of
course ﬁnite, and therefore the set J ′ ⊂ J of indices j such that xij ̸= 0 for some
⊕
1 ≤ i ≤ n is ﬁnite. It follows that ⟨X ⟩ ⊂ j ∈J ′ Mj M , contradiction.
Lemma 193. Let R1 , . . . , Rn be ﬁnitely many not necessarily commutative rings,
∏n
and put R = i=1 Ri . Then R is semisimple iﬀ Ri is semisimple for all 1 ≤ i ≤ n.
Exercise: Prove Lemma 193.
We now quote the following basic result from noncommutative algebra.
Theorem 194. (WedderburnArtin) For a ring R, TFAE:
(i) R is semisimple as a left Rmodule (left semisimple).
(ii) R is semisimple as a right Rmodule (right semisimple).
(iii) There are N, n1 , . . . , nN ∈ Z+ and division rings D1 , . . . , DN such that
R∼
= N
∏
i=1 Mni (Di ). 126 PETE L. CLARK Combining Theorems 191 and 194 gives us a tremendous amount of information.
First of all, a ring is left semisimple iﬀ it is right semisimple, so we may as well
speak of semisimple rings. A ring is semisimple iﬀ it is absolutely projective
iﬀ it is absolutely injective.
Coming back to the commutative case, the WedderburnArtin theorem tells us
that the class of semisimple / absolutely projective / absolutely injective rings is
extremely restricted.
Corollary 195. A commutative ring is semisimple iﬀ it is a ﬁnite product of ﬁelds.
However it is signiﬁcantly easier to give a proof of WedderburnArtin in the commutative case, so we will give a direct proof of Corollary 195
Proof. Step 1: Oﬃcially speaking the theorem holds for the zero ring because it is
an empty product of ﬁelds. In any event, we may and shall assume henceforth that
our semisimple ring is nonzero.
Step 0: A ﬁeld is a semisimple ring: e.g. every module over a ﬁeld is free, hence
projective. By Lemma 193, a ﬁnite direct product of ﬁelds is therefore semisimple.
⊕
Step 1: Let R be a semisimple ring, and let R = i∈I Mi be a direct sum decomposition into simple Rmodules. R is a ﬁnitely generated Rmodule, by Lemma 192 I
is ﬁnite, and we may identify it with {1, . . . , n} for some n ∈ Z+ : R = M1 ⊕ . . . ⊕ Mn .
Step 2: We may uniquely write 1 = e1 + . . . + en with ei ∈ Mi . Then for all i ̸= j ,
ei ej = 0, and this together with the identity 1 · 1 = 1 implies that e2 = ei for all
i
i. As usual for idempotent decompositions, this expresses R as a direct product of
the subrings Ri = Mi = ei R. Moreover, since Mi is a simple Rmodule, Ri has no
proper nonzero ideals, and thus it is a ﬁeld, say ki .
Exercise 8.7: Exhibit an absolutely ﬂat commutative ring which is not semisimple.
8.4. Normal Series.
If M is an Rmodule a normal series is a ﬁnite ascending chain of Rsubmodules
0 = M0 M1 . . . Mn = M . We say that n is the length of the series. (The
terminology is borrowed from group theory, in which one wants a ﬁnite ascending
chain of subgroups with each normal in the next. Of course there is no notion of
“normal submodule”, but we keep the grouptheoretic terminology.)
There is an evident partial ordering on the set of normal series of a ﬁxed Rmodule
′
′
M : one normal series {Mi }n is less than another normal series {Mj }n=0 if for all
i=0
j
′
1 ≤ i ≤ n, Mi is equal to Mj for some (necessarily unique) j . Rather than saying
′
′
that {Mi } ≤ {Mj }, it is traditional to say that the larger series {Mj } reﬁnes the
smaller series {Mi }.
Given any normal series {Mi } we may form the associated factor sequence
′
′
M1 /M0 = M1 , M2 /M1 , . . . , Mn /Mn−1 = M/Mn−1 . Two normal series {Mi }n , {Mj }n=0
i=0
j
are equivalent if n = n′ and there is a permutation σ of {1, . . . , n} such that for all
′
′
1 ≤ i ≤ n, the factors Mi /Mi−1 and Mσ(i) /Mσ(i)−1 are isomorphic. In other words,
if we think of the factor sequence of a normal series as a multiset of isomorphism
classes of modules, then two normal series are equivalent if the associated multisets
of factors are equal. COMMUTATIVE ALGEBRA 127 Exercise 8.8: Show that reﬁnement descends to a partial ordering on equivalence
classes of normal series of a ﬁxed Rmodule M .
The following theorem is the basic result in this area.
Theorem 196. (Schreier Reﬁnement) For any Rmodule M , the partially ordered
set of equivalence classes of normal series of submodules of M is directed: that is,
any two normal series admit equivalent reﬁnements.
Proof. For a proof in a context which simultaneously generalizes that of modules
and groups, see e.g. [J2, p. 106].
For an Rmodule M , a composition series is a maximal element in the poset of
normal series: that is, a composition series which admits no proper reﬁnement.
Exercise 8.9: Show that a normal series {Mi }n for an Rmodule M is a comi=0
position series if and only if for all 1 ≤ i ≤ n, the factor module Mi /Mi−1 is
simple.
o
Theorem 197. (JordanH¨lder) Let M be an Rmodule. Then any two composition series for M are equivalent: up to a permutation, their associated factor series
are termbyterm isomorphic.
Proof. This is an immediate consequence of Schreier Reﬁnement: any two normal
series admit equivalent reﬁnements, but no composition series admits a proper
reﬁnement, so any two composition series must already be equivalent!
Remark: The ordertheoretic core of the proof is simply that a directed set admits
at most one maximal element.
Thus for a module M which admits a composition series, we may deﬁne the length
ℓ(M ) of M to be the length of any composition series. One also speaks of the
JordanH¨lder factors of M or the composition factors of M, i.e., the unique
o
multiset of isomorphism classes of simple Rmodules which must appear as the successive quotients of any composition series for M .
If a module does not admit a composition series, we say that it has inﬁnite length.
And now a basic question: which Rmodules admit a composition series?
Exercise 8.10: a) Show that any ﬁnite35 module admits a composition series.
b) Show that if a module M admits a composition series, it is ﬁnitely generated.
c) Show that a Zmodule M admits a composition series iﬀ it is ﬁnite.
d) Let k be a ﬁeld. Show that a k module admits a composition series iﬀ it is
ﬁnitely generated (i.e., iﬀ it is ﬁnitedimensional).
Exercise 8.11: An Rmodule M admits a composition series iﬀ there exists L ∈ Z+
such that every normal series in M has length at most L.
35Recall that in these notes by a “ﬁnite module” we mean a module whose underlying set is
ﬁnite! 128 PETE L. CLARK Theorem 198. For an Rmodule M , TFAE:
(i) M is both Noetherian and Artinian.
(ii) M admits a composition series.
Proof. Assume (i). Since M satisﬁes (DCC), there must exist a minimal nonzero
submodule, say M1 . If M1 is a maximal proper submodule, we have a composition
series. Otherwise among all proper Rsubmodules strictly containing M1 , by (DCC)
we can choose a minimal one M2 . We continue in this way: since M also satisﬁes
(ACC) the process must eventually terminate, yielding a composition series.
(ii) =⇒ (i): This follows easily from Exercise 8.11.
Exercise 8.12: Exercise 8.11 (and hence also our proof of Theorem 198) makes use
of Schreier Reﬁnement. Give a proof that (ii) =⇒ (i) in Theorem 198 which is
independent of Schreier Reﬁnement. (Suggestion: try induction on the length of a
composition series.)
Proposition 199. Let 0 → M ′ → M → M ′′ → 0 be a short exact sequence of
Rmodules. Then:
a) M admits a composition series iﬀ both M ′ and M ′′ admit composition series.
b) If M admits a composition series, then
ℓ(M ) = ℓ(M ′ ) + ℓ(M ′′ ).
Exercise 8.13: Prove Proposition 199.
Remark: Although it will not play a prominent role in our course, the length of an
Rmodule M is an extremely important invariant, especially in algebraic geometry:
it is is used, among other things, to keep track of intersection multiplicities and to
quantitatively measure the degree of singularity of a point.
8.5. The KrullSchmidt Theorem.
The material in this section follows [J2, §3.4] very closely. In particular, very
exceptionally for us – but as in loc. cit. – in this section we work with left modules
over a possibly noncommutative ring R. The reason: not only does the desired
result carry over verbatim to the noncommutative case (this is not in itself a good
enough reason, as the same holds for a positive proportion of the results in these
notes) but the proof requires us to consider noncommutative rings!
A module M is decomposable if there are nonzero submodules M1 , M2 ⊂ M
such that M = M1 ⊕ M2 ; otherwise M is indecomposable.
Theorem 200. (KrullSchmidt) Let M be an Rmodule of ﬁnite length. Then:
⊕m
a) There are indecomposable submodules M1 , . . . , Mm such that M = i=1 Mi .
⊕n
b) If there are indecomposable submodules N1 , . . . , Nn such that M = i=1 Ni , then
m = n and there exists a bijection σ of {1, . . . , n} such that for all i, Mi ∼ Nσ(i) .
=
The Proof of Theorem 200a) is easy, and we give it now. If M is a ﬁnite length
module and we write M = M1 ⊕ M2 then 0 < ℓ(M1 ), ℓ(M2 ) < ℓ(M ). Thus an
evident induction argument shows that any sequence of moves, each one of which
splits a direct summand of M into two nontrivial direct subsummands of M , must
terminate after ﬁnitely many steps, leaving us with a decomposition of M into a
ﬁnite direct sum of indecomposable submodules. COMMUTATIVE ALGEBRA 129 As one might suspect, the second part of Theorem 200 concerning the uniqueness of the indecomposable decomposition is more subtle. Indeed, before giving the
proof we need some preparatory considerations on endomorphism rings of modules.
Proposition 201. For an Rmodule M , TFAE:
(i) M is decomposable.
(ii) The (possibly noncommutative, even if R is commutative) ring EndR (M ) =
HomR (M, M ) has a nontrivial idempotent, i.e., an element e ̸= 0, 1 with e2 = e.
Exercise 8.14: Prove Proposition 201.
A (not necessarily commutative) ring R is local if the set of nonunits R \ R×
forms a twosided ideal of R.
Exercise 8.15: Let R be a local, not necessarily commutative ring.
a) Show that R ̸= 0.
b) Show that R has no nontrivial idempotents.
An Rmodule M is strongly indecomposable if EndR (M ) is local. Thus it
follows from Proposition 201 and Exercise 8.15 that a strongly indecomposable
module is indecomposable.
Example: The Zmodule Z is indecomposable: any two nonzero submodules (a)
and (b) have a nontrivial intersection (ab). On the other hand EndZ (Z) = Z is not
a local ring, so Z is not strongly indecomposable.
Thus “strongly indecomposable” is, in general, a stronger concept than merely
“indecomposable”. Notice though that the KrullSchmidt theorem applies only to
ﬁnite length modules – equivalently to modules which are both Noetherian and
Artinian – and Z is not an Artinian Zmodule. In fact, it shall turn out that any
ﬁnite length indecomposable module is strongly indecomposable, and this will be a
major step towards the proof of the KrullSchmidt Theorem.
But we are not quite ready to prove this either! First some Fitting theory.
For an Rmodule M and f ∈ EndR (M ), we put
f ∞ (M ) = ∞
∩ f n (M ). n=1
∞ The set f (M ) is the intersection of a descending chain
M ⊃ f (M ) ⊃ f 2 (M ) ⊃ . . . ⊃ f n (M ) ⊃ . . .
of submodules of M , and is thus an f stable submodule of M . The restriction of f
to f ∞ (M ) is surjective. Moreover, if M is an Artinian module, there exists s ∈ Z+
such that f s (M ) = f s+1 (M ) = . . ..
Exercise 8.16: Find a commutative ring R, an Rmodule M and f ∈ EndR (M )
such that for no n ∈ Z+ is the submodule f n (M ) f stable. 130 PETE L. CLARK Similarly, for M and f as above, we put
f−∞ (0) = ∞
∪ ker f n . n=1 Here each ker f n is an f stable submodule of M on which f is nilpotent. The set
f−∞ (0) is the union of an ascending chain of submodules
0 ⊂ ker f ⊂ ker f 2 ⊂ . . . ⊂ ker f n ⊂ . . .
of M and is thus an f stable submodule of M on which f acts as a nil endomorphism: i.e., every element of M is killed by some power of f . Moreover, if M is a
Noetherian module, there exists t ∈ Z+ such that ker f t = ker f t+1 = . . . and thus
f is a nilpotent endomorphism of f−∞ (0).
Exercise 8.17: Find a commutative ring R, an Rmodule M and f ∈ EndR (M )
such that f is not a nilpotent endomorphism of f−∞ (0).
Theorem 202. (Fitting’s Lemma) Let M be a ﬁnite length module over the not
necessarily commutative ring R, and let f ∈ EndR (M ).
a) There exists a Fitting Decomposition
(9) M = f ∞ (M ) ⊕ f −∞ (0). b) f f ∞ (M ) is an isomorphism and f f−∞ (0) is nilpotent.
Proof. Since M has ﬁnite length it is both Noetherian and Artinian. Thus there
exists r ∈ Z+ such that
f r (M ) = f r+1 (M ) = . . . = f ∞ (M )
and
ker f r = ker f r+1 = . . . = f−∞ (0).
Let x ∈ f ∞ (M ) ∩ f−∞ (0). Then there exists y ∈ M such that x = f r (y ); moreover
0 = f r (x) = f 2r (y ). But f 2r (y ) = 0 implies x = f r (y ) = 0, so f ∞ (M )∩f−∞ (0) = 0.
Let x ∈ M . Then f r (x) ∈ f r (M ) = f 2r (M ), so there exists y ∈ M with
r
f (x) = f 2r (y ) and thus f r (x − f r (y )) = 0. so
x = f r (y ) + (x − f r (y )) ∈ f ∞ (M ) + f −∞ (0),
completing the proof of part a). As for part b), we saw above that the restriction of
f to f ∞ (M ) is surjective. It must also be injective since every element of the kernel
lies in f −∞ (0). Thus f f ∞ (M ) is an isomorphism. Finally, as observed above, since
f−∞ (0) = ker f r , f f−∞ (0) is nilpotent.
Lemma 203. Let x and y be nilpotent elements in a not necessarily commutative
ring (which is not the zero ring). Then x + y is not a unit of R.
Proof. Assume to the contrary that x + y = u ∈ R× . Dividing through by u we
reduce to showing that we cannot have two nilpotent elements∑ y ∈ R such that
x,
∞
x + y = 1. But if x is nilpotent, the “inﬁnite geometric series” n=0 1 + x + . . . +
xn + . . . is in fact ﬁnite, hence perfectly legal in our abstract algebraic context, and
it is immediate to check that the familiar calculus identity
∞
∑
(1 − x)(
xn ) = 1
n=0 COMMUTATIVE ALGEBRA 131 holds here. Thus y = 1−x is both a unit of R and a nilpotent element, contradiction.
Corollary 204. Let M be a ﬁnite length indecomposable Rmodule. Then every
f ∈ EndR (M ) is either an automorphism or nilpotent. Moreover M is is strongly
indecomposable.
Proof. Since M is indecomposable, Fitting’s Lemma implies that for f ∈ EndR (M )
we must have either M = f ∞ (M ) – in which case f is an automorphism – or
M = f −∞ (0) – in which case f is nilpotent. We must show that I = EndR (M ) \
EndR (M )× is a twosided ideal of EndR (M ). Note that I is precisely the set of
endomorphisms of M which are not automorphisms, hence every element of I is
nilpotent. By Lemma 203, I is a subgroup of (R, +). Moreover, for f ∈ I, g ∈
EndR (M ), since f is neither injective nor surjective, gf is not injective and f g is
not surjective, so neither is an automorphism and both lie in I .
Lemma 205. Let M be a nonzero Rmodule and N an indecomposable Rmodule.
Suppose we have homomorphisms f : M → N, g : N → M such that gf is an
automorphism of M . Then both f and g are isomorphisms.
Proof. Let h = (gf )−1 , l = hg : N → M and e = f l : N → N . Then lf = hgf =
1M and e2 = f lf l = f 1M l = f l = e. Since M is indecomposable, either e = 1
or e = 0, and the latter implies 1M = 12 = lf lf = lef = 0, i.e., M = 0. So
M
f l = e = 1N , so f is an isomorphism and thus so too is (f (gf )−1 )−1 = g .
⊕m
Theorem 206. Let M ∼ N be isomorphic modules, and let M =
=
i=1 Mi and
⊕n
N = i=1 Ni′ with each Mi strongly indecomposable and each Ni indecomposable.
Then m = n and there is a bijection σ of {1, . . . , m} such that for all i, Mi ∼ Nσ(i) .
=
Proof. By induction on m: m = 1 is clear. Suppose the result holds for all direct
sums of fewer than m strongly indecomposable submodules.
Step 1: Let e1 , . . . , em ∈ EndR (M ) and f1 , . . . , fn ∈ EndR (N ) be the idempotent
elements corresponding to the given direct sum decompositions (i.e., projection
∼
onto the corresponding factor). Let g : M → N , and put
hj := fj ge1 ∈ HomR (M, N ), kj = e1 g −1 fj ∈ HomR (N, M ), 1 ≤ j ≤ n.
Then
n
∑
j =1 kj hj = ∑ e1 g −1 fj ge1 = e1 g −1 j ∑ fj ge1 = e1 g −1 1N ge1 = e1 . j The restrictions of e1 and kj hj to M1 stabilize M1 so may be regarded as endomorphisms of M1 , say e′ and (kj hj )′ , and we have
1
n
∑ (kj hj )′ = e′ = 1M1 .
1 j =1 By assumption EndR M1 is local, so for at least one j , (kj hj )′ is a unit, i.e., an
automorphism of M1 . By reordering the Nj ’s we may assume that j = 1, so
(k1 h1 )′ ∈ AutR M1 . We may regard the restriction h′ of h1 to M1 as a homomor1
′
phism from M1 to N1 and similarly the restriction k1 of k1 to N1 as a homomor′′
′
phism from N1 to M1 , and then k1 h1 = (k1 h1 ) is an automorphism. By Lemma 132 PETE L. CLARK
∼ ∼ ′
205, h′ = (f1 ge′ ) : M1 → N1 and k1 = (e1 g −1 f1 )′ : N1 → M1 .
1
1
Step 2: We claim that
m
⊕
Mi .
(10)
M = g −1 (N1 ) ⊕
i=2 ⊕m
To see this, let x ∈ g N1 ∩ ( i=2 Mi ), so x = g −1 y for some y ∈ N1 . Because
⊕m
x ∈ i=2 Mi , e1 x = 0. Thus
−1 ′
0 = e1 x = e1 g −1 y = e1 g −1 f1 y = k1 y = k1 y.
′
Since k1 is an isomorphism, y = 0 and thus x = 0, so the sum in (10) is direct.
⊕m
Now put M ′ = g −1 (N1 ) ⊕ i=2 Mi , so we wish to show M ′ = M . Let x ∈ g −1 N1 .
Then x, e2 x, . . . , em x ∈ M ′ , so e1 x = (1 − e2 − . . . − em )x ∈ M ′ . So
′
M ′ ⊃ e1 g −1 N1 = e1 g −1 f1 N1 = k1 N1 = k1 N1 = M1
⊕m
and thus M ′ ⊃ i=1 Mi = M .
∼
Step 3: The isomorphism g : M → N carries g −1 N1 onto N1 hence induces an
sim
isomorphism g−MN1 → N/N1 . Using Step 2, we have
1
n
⊕
j =2 N ∼ M ∼⊕
Mi .
= −1
=
N1
g N1
i=2
m Ni = We are done by induction.
Exercise 8.18: Please conﬁrm that we have proved the KrullSchmidt Theorem!
Exercise 8.19: Let M and N be Rmodules such that M × M ∼ N × N .
=
a) If M and N are both of ﬁnite length, show that M ∼ N .
=
b) Must we have M ∼ N in general?
=
Remark: Part b) is far from easy! If you give up, see [Cor64].
Exercise 8.20: a) Let R be a PID and M an Rmodule. a) Show that M has
ﬁnite length iﬀ it is a ﬁnitely generated torsion module.
b) Show that a ﬁnitely generated torsion module is indecomposable iﬀ it is isomorphic to R/(pa ) for some prime element p of R and some a ∈ Z+ .
c) Did you use the structure theorem for ﬁnitely generated modules over a PID to
prove parts a) and b)? If so, try to prove these results without it.
d) Take as given parts a) and b) of this exercise, and use the KrullSchmidt Theorem to deduce the structure theorem for ﬁnitely generated modules over a PID.
Remark: Later we will use these ideas to give an independent proof of the structure
theorem for ﬁnitely generated modules over a PID, using one extra idea: reduction
to the case of a local PID, in which case there is only one nonzero prime ideal and
the module theory becomes especially simple.
8.6. Some important terminology.
All we aspire to do in this section is to introduce some terminology, but it is so
important that we have isolated it for future reference. COMMUTATIVE ALGEBRA 133 Let R be a ring and p a prime ideal of R. The height of p is the supremum
of all lengths of ﬁnite chains of prime ideals of the form p0
p1
...
pn = p
(the length of the indicated chain being n; i.e., it is the number of ’s appearing, which is one less than the number of elements). Thus the height is either a
nonnegative integer or ∞; the latter transpires iﬀ there exist arbitrarily long ﬁnite
chains of prime ideals descending from p (and of course, this need not imply the
existence of an inﬁnite chain of prime ideals descending from p).
A prime ideal of height 0 is called a minimal prime. In an integral domain
R, the unique minimal prime is (0), so the concept is of interest only for rings
which are not domains. If I is a proper ideal of R, we also speak of a minimal
prime over I , which means a prime p ⊃ I such that there is no prime ideal q with
I
q p. Note that p is a minimal prime over I iﬀ p is a minimal prime in the
quotient ring R/I . This remark simultaneously explains the terminology “minimal
over” and gives a hint why it is useful to study minimal prime ideals even if one is
ultimately most interested in integral domains.
The dimension of a ring R is the supremum of all the heights of its prime ideals.
The full proper name here is Krull dimension of R, which is of course useful
when one has other notions of dimension at hand. Such things certainly do exist
but will not be considered here. Moreover, as will shortly become apparent, the
need to include Krull’s name here so as to ensure that he gets proper recognition
for his seminal work in this area is less than pressing. Therefore we use the full
name “Krull dimension” only rarely as a sort of rhetorical ﬂourish.
One also often speaks of the codimension of a prime ideal p of R, which is the
dimension of R minus the height of p. This is especially natural in applications
to algebraic geometry, of which the present notes allude to only in passing. Note
that this is not necessarily equal to the Krull dimension of R/p – or what is the
same as that, the maximal length of a ﬁnite chain of prime ideals ascending from p
– although in reasonable applications, and especially in geometry, one is certainly
entitled to hope (and often, to prove) that this is the case.
Remark: All of these deﬁnitions would make perfect sense for arbitrary partially
ordered sets and their elements, but the terminology is not completely consistent
with order theory. Namely, the height of an element in an arbitrary poset is deﬁned as the supremum of lengths of chains descending from that element, but the
order theorists would cringe to hear the supremum of all heights of elements called
the “dimension” of the poset. They would call that quantity the height of the
poset, and would reserve dimension for any of several more interesting invariants.
(Roughly, the idea is that a chain of any ﬁnite length is onedimensional, whereas
a product of d chains should have dimension d.)
8.7. Introducing Noetherian rings.
The following is arguably the most important single deﬁnition in all of ring theory.
A ring R is said to be Noetherian if the poset I (R) of all ideals of R satisﬁes
the ascending chain condition. 134 PETE L. CLARK Exercise 8.21: a) Show that a ﬁnite ring is Noetherian.
b) For each cardinal κ ≥ ℵ0 , exhibit a ring of cardinality κ which is not Noetherian.
c) A ring R is said to have ﬁnite quotients if for any 0 ̸= I in R, R/I is ﬁnite.
Show that a ring with ﬁnite quotients is Noetherian.
d) Deduce that Z is Noetherian.
Theorem 207. A ﬁnitely generated module over a Noetherian ring is Noetherian.
Proof. If M is a ﬁnitely generated module over R, then we may represent it as
Rn /K for some submodule K of Rn . An immediate corollary of the preceding
theorem is that ﬁnite direct sums of Noetherian modules are Noetherian, and by
assumption R itself is a Noetherian Rmodule, hence so is Rn and hence (again by
Theorem XX) so is the quotient Rn /K = M .
Thus so long as we restrict to Noetherian rings, submodules of ﬁnitely generated
modules remain ﬁnitely generated. This is extremely useful even in the case of
R = Z: a subgroup of a ﬁnitely generated abelian group remains ﬁnitely generated.
Needless(?) to say, this does not hold for all nonabelian groups, e.g. not for a
ﬁnitely generated free group of rank greater than 1.
Theorem 208. (Characterization of Noetherian rings) For a ring R, TFAE:
(i) Every nonempty set of ideals of R has a maximal element.
(ii) There are no inﬁnite ascending chains
I1 I2 ... In ... of ideals of R.
(iii) Every ideal of R is ﬁnitely generated.
(iv) Every prime ideal of R is ﬁnitely generated.
Proof. Exercise X.X displays the equivalence of (i) and (ii). Assuming (ii), let I be
an ideal of R. If I cannot be ﬁnitely generated, this means that we can choose a
sequence of elements {xi } of I such that
0 ⟨x1 ⟩ ⟨x1 , x2 ⟩ ..., contradicting (ACC). Assuming (iii), considering an inﬁnite ascending chain of
∪
ideals as in (ii) above, put J = i Ii . Then J is ﬁnitely generated, say by elements
x1 ∈ Ii1 , . . . , xr ∈ Iir . But then taking i to be the largest of the indices i1 , . . . , ir
we have
⟨x1 , . . . , xn ⟩ ⊂ Ii ⊂ J = ⟨x1 , . . . , xr ⟩,
so J = Ii = Ii+1 = . . . and the chain stabilizes at I .
Trivially (iii) =⇒ (iv). That (iv) implies (iii) is the most diﬃcult implication,
and is another instance of the metaprinciple that ideals which are maximal with
respect to some property tend to be prime. Indeed, this will be a consequence of
the following result.
Lemma 209. Let R be a ring admitting at least one nonﬁnitely generated ideal.
a) Then there exists an ideal which is maximal among all ideals which are not
ﬁnitely generated.
b) Any ideal I which is maximal among all ideals in R which are not ﬁnitely generated is necessarily prime. COMMUTATIVE ALGEBRA 135 Proof. For part a) we need only observe that the union of a chain of nonﬁnitely
generated ideals remains nonﬁnitely generated for similar reasons to the above:
a ﬁnite generating set for the union would be a ﬁnite generating set for some
suﬃciently large element of the chain. So suppose I is an ideal which is maximal
with respect to not being ﬁnitely generated, and let x, y ∈ R be such that xy ∈ I .
Assume for a contradiction that neither x nor y is in I . Then the ideal I ′ = ⟨I, a⟩
is strictly larger and thus ﬁnitely generated: say
I ′ = ⟨i1 + a1 x, . . . , in + an x⟩.
Now let J be the set of all b ∈ R with bx ∈ I . Then J contains both I and y , so
is strictly larger than I and thus ﬁnitely generated. We claim I = ⟨i1 , . . . , in , Jx⟩.
Indeed, take any z ∈ I . Then z ∈ I ′ , so
x = w1 (i1 + a1 x) + . . . + wn (in + an x).
Now u1 x1 + . . . + un xn ∈ J , so z ∈ ⟨i1 , . . . , in , Ja⟩. Thus the claim is established,
and, since J is ﬁnitely generated, so is Ja and thus I , contradiction.
Exercise 8.22: Suppose a ring R satisﬁes the ascending chain condition on prime
ideals. Must R be Noetherian?
Proposition 210. Let R be a Noetherian ring.
a) If I is any ideal of R, the quotient R/I is Noetherian.
b) If S ⊂ R is any multiplicative subset, the localization S −1 R is Noetherian.
Proof. Any ideal of R/I is of the form J/I for some ideal J ⊃ I of R. By assumption
J is ﬁnitely generated, hence J/I is ﬁnitely generated, so R/I is Noetherian. A
similar argument holds for the localization; details are left to the reader.
∏
Exercise 8.23: Let k be a ﬁeld, let S be an inﬁnite set, and put R = s∈S k , i.e.,
the inﬁnite product of #S copies of k . Show that R is not Noetherian, but the
localization Rp at each prime ideal is Noetherian.
Thus Noetherianity is a localizable property but not a local property.
In each of the next three subsections we will state and prove an extremely important and general theorem about Noetherian rings.
8.8. The BassPapp Theorem.
We now present a beautiful characterization of Noetherian rings in terms of properties of injective modules, due independently to Z. Papp [Pa59] and H. Bass [Bas59].
Theorem 211. (BassPapp Theorem) For a ring R, TFAE:
(i) A direct limit of injective modules is injective.
(ii) A direct sum of injective modules is injective.
(iii) A countable direct sum of injective modules is injective.
(iv) R is Noetherian.
Proof.
(i) =⇒ (ii): A direct sum is a kind of direct limit. (ii) =⇒ (iii) is immediate. 136 PETE L. CLARK (iii) =⇒ (iv): Let I1 ⊂ I2 ⊂ . . . ⊂ In ⊂ . . . be an inﬁnite ascending chain of ideals
∪
of R, and let I = n In . We deﬁne
E= ∞
⊕ E (R/In ). n=1 For n ∈ Z+ , let fn : I → E (R/In ) be the composite map I → R → R/In →
∏
∏∞
E (R/In ). There is then a unique map f : I → n=1 E (R/In ). But indeed, for
each ∏
ﬁxed x ∈ I , x lies in In for suﬃciently large n and thus fn (x) = 0. It follows
that f actually lands in the direct sum, and we have thus deﬁned a map
f : I → E.
By hypothesis, E is a countable direct sum of injective modules and therefore
injective, so f extends to an Rmodule map with domain all of R and is thus of the
form f (x) = xf (1) = xe for some ﬁxed e ∈ E . Let N be suﬃciently large so that
for n ≥ N , the nth component en of e is zero. Then for all x ∈ I ,
0 = xen = fn (x) = x + In ∈ R/In ,
and thus x ∈ In . That is, for all n ≥ N , In = I .
(iv) =⇒ (i): let {Eα } be a directed system of injective modules with direct limit
E . For α ≤ β we denote the transition map from Eα to Eβ by ιαβ and the natural
map from Eα to E by ια . We will show E is injective by Baer’s Criterion (Theorem
29), so let I be any ideal of R and consider an Rmodule map f : I → E . Since
R is Noetherian, I is ﬁnitely generated, and it follows that there exists an index
α such that f (I ) ⊂ ια (Eα ). Let M be a ﬁnitely generated submodule of Eα such
that f (I ) ⊂ ια (M ). Consider the short exact sequence
ι α
0 → K → M → f (I ) → 0. Since M is ﬁnitely generated and R is Noetherian, K is ﬁnitely generated. Moreover
K maps to 0 in the direct limit, so there exists β ≥ α such that ιαβ K = 0. Let
M ′ = ιαβ M , so by construction
∼ ιβ : M ′ → f (I ).
Taking g = ιβ −1′ ◦ f we get a map g : I → Eβ such that f = ιβ ◦ g . Since Eβ is
M
injective, g extends to a map G : R → Eβ and thus F = ιβ ◦ G extends f to R.
8.9. Artinian rings: structure theory.
A ring R which satisﬁes the descending chain condition (DCC) on ideals is called
Artinian (or sometimes, “an Artin ring”).
Exercise 8.24
a) Show that a ring with only ﬁnitely many ideals is Artinian.
b) Show that the ring of integers Z is not Artinian.
c) Show that a quotient of an Artinian ring is Artinian.
d) Show that a localization of an Artinian ring is Artinian.
Obviously any ﬁnite ring has only ﬁnitely many ideals and is Artinian. It is not
diﬃcult to give examples of inﬁnite rings with ﬁnitely many ideals. For instance,
let k be a ﬁeld and let 0 ̸= f ∈ k [t]. Then R = k [t]/(f ) has only ﬁnitely many COMMUTATIVE ALGEBRA 137 a
a
ideals. Indeed, if we factor f = f1 1 · · · fr r into irreducible factors, then the Chinese
Remainder Theorem gives
k [t]/(f ) ∼ k [t]/(f a1 ) × . . . × k [t]/(f ar ).
=
r 1 Each factor ring
are precisely k [t]/(fiai ) is a local ring with maximal ideal (f1 ), and the ideals
(0) (fi )ai −1 ... fi . Since every ideal in a product is a direct sum of ideals of the factors, there are then
∏r
precisely i=1 (ai + 1) ideals of R.
A bit of reﬂection reveals that – notwithstanding their very similar deﬁnitions –
requiring (DCC) on ideals of a ring is considerably more restrictive than the (ACC)
condition. For instance:
Proposition 212. A domain R is Artinian iﬀ it is a ﬁeld.
Proof. Obviously a ﬁeld satisﬁes (DCC) on ideals. Conversely, if R is a domain
and not a ﬁeld, there exists a nonzero nonunit element a, and then we have (a)
(a2 ) (a3 ) . . .. Indeed, if (ak ) = (al ), suppose k ≤ l and write l = k + n, and
then we have uak = ak an for some u ∈ A× , and then by cancellation we get an = u,
so an is unit and thus a is a unit, contradiction.
The result collects several simple but important properties of Artinian rings.
Theorem 213. Let R be an Artinian ring.
a) R has dimension zero: prime ideals are maximal.
b) Therefore the Jacobson radical of R coincides with its nilradical.
c) R has only∩
ﬁnitely many maximal ideals, say m1 , . . . , mn .
n
d) Let N = i=1 mi be the nilradical. Then it is a nilpotent ideal: there exists
+
k ∈ Z such that N k = 0.
Proof. a) If p is a prime ideal of A, then A/p is an Artinian domain, which by
Proposition XX is a ﬁeld, so p is maximal.
b) By deﬁnition, the Jacobson radical is the intersection of all maximal ideals and
the nilradical is the intersection of all prime ideals. Thus the result is immediate
from paqrt a).
c) Suppose mi is an inﬁnite sequence of maximal ideals. Then
R m1 m1 ∩ m2 · · · is an inﬁnite descending chain. Indeed, equality at any step would mean mN +1 ⊃
∏N
∩N
i=1 mi , and then since mN +1 is prime it contains mi for some 1 ≤ i ≤
i=1 mi =
N , contradiction.
d) Applying DCC on the powers of N , it must be the case that there exists some
k with N k = N k+n for all n ∈ Z+ . Put I = N k . Suppose I ̸= 0, and let Σ be
the set of ideals J such that IJ ̸= 0. Evidently Σ ̸= ∅, for I ∈ Σ. By DCC we are
entitled to a minimal element J of σ . There exists 0 ̸= x ∈ J such that xI ̸= 0.
For such an x, we have (x) ∈ Σ and by minimality we must have J = (x). But
(xI )I = xI ̸= 0, so xI ⊂ (x) and thus xI = (x) by minimality. So there exists
y ∈ I with xy = x and thus we have
x = xy = xy 2 = . . . = xy k = . . . . 138 PETE L. CLARK But y ∈ I ⊂ N , hence y is nilpotent and the above equations give x = 0, a
contradiction.
Lemma 214. Suppose that in a ring R there exists a ∏
ﬁnite sequence m1 , . . . , mn of
(not necessarily distinct) maximal ideals such that 0 = i mi . Then R is Noetherian
iﬀ it is Artinian.
Proof. Consider the chain of ideals
R ⊃ m1 ⊃ m1 m2 ⊃ . . . ⊃ ∏ mi = 0. i Each quotient Qi := m1 · · · mi−1 /m1 · · · mi is a vector space over R/mi . Now R
satisﬁes (ACC) (resp. (DCC)) for ideals iﬀ each Qi satisﬁes (ACC) (resp. (DCC)).
But since each Qi is a vector space, (ACC) holds iﬀ (DCC) holds (iﬀ the dimension
is ﬁnite).
Theorem 215. (AkizukiHopkins) For a ring R, TFAE:
(i) R is Artinian.
(ii) R is Noetherian, and prime ideals are maximal.
Proof. Assume R is Artinian. Then by Proposition X.X prime ideals are maximal,
so it suﬃces to show that R is Noetherian. Let m1 , . . . , mn be the distinct maximal
∏n
∩n
k
ideals of R. For any k ∈ Z+ we have i=1 mk ⊂ ( i=1 mi ) . Applying Theorem
i
∏n
213d), this shows that for suﬃciently large k we have i=1 mk = 0. We can now
i
apply Lemma 214 to conclude that R is Artinian. Conversely, suppose R is Noetherian and zerodimensional. Like any Noetherian ring, R has only ﬁnitely many
minimal prime ideals, each of which is maximal by zerodimensionality. Therefore
∩n
N = i=1 mi is the nilradical of a Noetherian ring, hence a nilpotent ideal. As
∏n
above, we deduce that for suﬃciently large k we have i=1 mk = 0 and then we
i
apply Lemma 214 again to conclude that R is Artinian.
Exercise 8.25: Let k be a ﬁeld and A = k [{xi }∞ ] a polynomial ring over k in
i=1
a countable inﬁnite number of indeterminates. Let m = ({xi }) be the ideal of all
polynomials with zero constant term, and put R = A/m2 . Show that R is a ring
with a unique prime ideal which is not Noetherian (so also not Artinian).
Exercise 8.26: Let n ∈ Z+ . Suppose R is a Noetherian domain with exactly n
prime ideals. Must R be Artinian?
Proposition 216. Let (R, m) be a Noetherian local ring.
a) Either:
(i) mk ̸= mk+1 for all k ∈ Z+ , or
(ii) mk = 0 for some k .
b) Moreover, condition (ii) holds iﬀ R is Artinian.
Proof. a) Suppose there exists k such that mk = mk+1 . By Nakayama’s Lemma,
we have mk = 0. If p is any prime ideal of R, then mk ⊂ p, and taking radicals we
have m ⊂ p, so p = m and R is a Noetherian ring with a unique prime ideal, hence
an Artinian local ring. b) If R is Artinian, then (i) cannot hold, so (ii) must hold.
Conversely, if (ii) holds then m is a nil ideal, hence contained in the intersection of
all prime ideals of R, which implies that m is the only prime ideal of R, and R is
Artinian by the AkizukiHopkins theorem. COMMUTATIVE ALGEBRA 139 Theorem 217. (Structure theorem for Artinian rings) Let R be an Artinian ring.
∏n
a) There exist ﬁnitely many local Artinian rings Ri such that R ∼ i=1 Ri .
=
∏m
b) Moreover, the decomposition is unique in the sense that if R ∼ j =1 Sj is
=
another decomposition, then n = m and there exists a permutation σ of {1, . . . , n}
such that Ri ∼ Sσ(i) for all i.
=
Proof. a) Let (mi )n be the distinct maximal ideals of R. We have seen that there
i=1 ∏
n
exists∩ ∈ Z+ such that i=1 mk = 0. By XX the ideals mk are pairwise comaximal,
k
i
i
∏k
so so i mk = i mi . Therefore by CRT the natural mapping
i
A→ n
∏A
mk
i
i=1 is an isomorphism. Evidently this is a ﬁnite decomposition into a product of local
Artinian rings, giving part a).
b) The proof requires primary decomposition, so must be deferred to §10.5.
Exercise: Let R be an Artinian ring.
a) Show that every element of R is either a unit or a zero divisor.
b) Show that R is its own total fraction ring.
Exercise:36 For a ring R, TFAE:
(i) R is semilocal [sic!], i.e., MaxSpec R is ﬁnite.
(ii) R/ rad R is Artnian.
8.10. The Hilbert Basis Theorem.
The following result shows in one fell swoop that the majority of the rings that
one encounters in classical algebraic geometry and number theory are Noetherian.
Theorem 218. (Hilbert Basis Theorem) Let R be a Noetherian ring. Then: a)
The univariate polynomial ring R[t] is Noetherian.
b) The formal power series ring R[[t]] is Noetherian.
Proof. Seeking a contradiction, suppose J is an ideal of R[t] which is not ﬁnitely
generated. We inductively construcct a sequence f0 , f1 , . . . , fn , . . . of elements of J
and a sequence of ideals Jn = ⟨f0 , . . . , fn ⟩ of R[t] as follows: f0 = 0, and for all
i ∈ N, fi+1 is an element of minimal degree in J \ Ji . Moreover, for all i ∈ Z+ let
ai be the leading coeﬃcient of fi , and let I be the ideal ⟨a1 , a2 , . . . , aN , . . .⟩ of R.
However, R is Noetherian, so there exists N ∈ Z+ such that I = ⟨a1 , . . . , aN ⟩. In
particular, there are u1 , . . . , uN ∈ R such that aN +1 = u1 a1 + . . . + uN aN . Dene
g= N
∑ ui fi tdeg fN +1 −deg fi . i=1 Since g ∈ JN and fN +1 ∈ J \ JN , fN +1 − g ∈ J \ JN . Moreover, by construction
g and fN +1 have the same degree and the same leading term, so deg fN +1 − g <
deg fN +1 , hence fN +1 does not have minimal degree among polynomials in J \ JN ,
contradiction.37
36This exercise should be compared to Exercise 4.14, which gives a criterion for semilocality
in terms of the quotient by the Jacobson radical.
37This argument is taken essentially verbatim from wikipedia. I like it a little better than the
standard writeups! 140 PETE L. CLARK b) Put S = R[[t]], and let I be an ideal of S . For r ≥ 0, put let Ii be the ideal of R
generated by the leading coeﬃcients ai of all elements f = ai ti + ai+1 ti+1 + . . . with
f ∈ I ∩ ti B . If f = ai ti + O(ti+1 ) ∈ Ii , then tf = ai ti+1 + O(ti+2 ) ∈ I ∩ ti+1 B , so
Ii ⊂ Ii+1 for all i. By the assumed Noetherianity of R, there must therefore be an N
such that IN = IN +k for all k ∈ N. For each 0 ≤ i ≤ N , let ai1 , . . . , aiM be a ﬁnite
set of generators for Ii 38, and let gij ∈ S be a power series whose leading coeﬃcient
is aij . We claim that I = ⟨gij ⟩1≤i≤N,1≤j ≤M . Indeed, let f ∈ I . By choosing a
suitable Alinear combination g0 of the g0j ’s we may arrange that f − g0 ∈ I ∩ tS ,
and similarly for 1 ≤ i ≤ N there exists gi ∈ Ii such that
f− N
∑ gi ∈ I ∩ tN +1 S. i=0 Moreover, IN +1 = IN , so there exists gN +1 , an Alinear combination of the tgN j ’s,
such that
N +1
∑
f−
gi ∈ I ∩ tN +2 S.
i=0 Continuing on in this way, we get an expression of f as an inﬁnite power series,
which can easily be regrouped to give an expression of the form
f = g0 + . . . + gN + M
∑ hj g N j , j =1 QED.
Exercise 8.27: Prove the converse of the Hilbert Basis Theorem: if R is a ring such
that either R[t] or R[[t]] is Noetherian, then R is Noetherian.
Corollary 219. A ﬁnitely generated algebra over a Noetherian ring is Noetherian.
∼
Proof. Let R be Noetherian and S a ﬁnitely generated Ralgebra, so that S =
R[t1 , . . . , tn ]/I for some n ∈ Z+ and some ideal I . By the Hilbert Basis Theorem
(and induction), R[t1 , . . . , tn ] is Noetherian, hence so is its quotient ring S .
Exercise 8.28: Show that for a ring R, TFAE:
(i) R is Noetherian.
(ii) For any n ≥ 1, R[[t1 , . . . , tn ]] is Noetherian.
Exercise 8.29: Let k be a ﬁeld, and consider the subring R = k [y, xy, x2 y, . . .]
of k [x, y ]. Show that R is not Noetherian.
Therefore, a subring of a Noetherian ring need not be Noetherian. Thinking that
this ought to be the case is one of the classic “rookie mistakes” in commutative
algebra. In general though, it is the exception rather than the rule that a nice
property of a ring R is inherited by all subrings of R, and one gets used to this.
8.11. The Krull Intersection Theorem.
38The number of generators M should depend on i, but since there are only ﬁnitely many ideals
in question, we may assume WLOG that each is generated by M generators for some M ∈ Z+ . COMMUTATIVE ALGEBRA 141 8.11.1. Preliminaries on Graded Rings.
In the proof of the theorem of this section we will need a little fact about homogeneous polynomials. So here we discuss some bare rudiments of this theory
by embedding it into its natural context: graded rings. The notion of graded
ring is of the utmost importance in various applications of algebra, from algebraic
geometry to algebraic topology and beyond. It would certainly be nice to give a
comprehensive exposition of graded algebra but at the moment this is beyond the
ambition of these notes, so we content ourselves with the bare minimum needed for
our work in the next section.
Let R be a ring, n ∈ Z+ , and denote by R[t] = R[t1 , . . . , tn ] the polynomial ring
in n indeterminates over R. For a polynomial P = P (t) in several variables, we
have the notion of the degree of P with respect to the variable ti : thinking of P
as an element of R[t1 , . . . , ti−1 , ti+1 , . . . , tn ][ti ] it is just the largest m such that the
coeﬃcient of tm is nonzero, as usual. For any monomial term cI ti1 · · · tin we deﬁne
n
i
1
the total degree to be d = i1 + . . . + in .
∑
A nonzero polynomial P = I cI ti1 · · · tin is homogeneous if all of its monomial
n
1
terms have the same total degree, and this common number is called the degree
of the homogeneous polynomial P . By convention the zero polynomial is regarded as being homogeneous total degree d for all d ∈ N.
A general polynomial P ∈ R[t] can be written as a sum of homogeneous poly∑∞
nomials P =
d=0 Pd (t) with each Pd homogeneous of degree d (and of course
Pd = 0 for all suﬃciently large d). This sum is unique. One way to see this is
to establish the following more structural fact: for any d ∈ N, let P [t]d be the
set of all polynomials which are homogeneous of degree d. Then each P [t]d is an
Rsubmodule of P [t] and we have a direct sum decomposition
(11) P [t ] = ∞
⊕ P [t ]d . d=0 Moreover, for all d1 , d2 ∈ N we have
(12) P [t]d1 · P [t]d2 ⊂ P [t]d1 +d2 . In general, if R is a ring and S is an algebra admitting an Rmodule direct sum
⊕∞
decomposition S = d=0 Sd satisfying Sd1 · Sd2 ⊂ Sd1 +d2 , then we say that S is an
(N)graded Ralgebra. Taking R = Z we get the notion of a graded ring.
⊕∞
Exercise: Let S = d=0 Sd be a graded Ralgebra. Show that the Rsubmodule
S0 is in fact an Ralgebra.
⊕∞
Let S =
d=0 Sd be a graded ring. We say that x ∈ S is homogeneous of
degree d if x ∈ Sd . An ideal I of S is homogeneous if it has a generating set
I = ⟨xi ⟩ with each xi a homogeneous element.
Exercise: Let S be a graded Ralgebra and let I be a homogeneous ideal of S . 142 PETE L. CLARK Show that
S/I = ∞
⊕
(Sd + I )/I
d=0 and thus S/I is a graded Ralgebra.
Back to the case of polynomial rings.
Lemma 220. Let S be a graded ring, let f1 , . . . , fn be homogeneous elements of S
and put I = ⟨f1 , . . . , f⟩ , so that I is a homogeneous ideal of S . Let f ∈ I be any
homogeneous element. Then there exist homogeneous elements g1 , . . . , g∈ R such
that
n
∑
f=
gi fi
i=1 and for all 1 ≤ i ≤ n deg gi = deg f − deg fi . Proof. Since f ∈ I , there exist X1 , . . . , Xn ∈ S such that
f = X1 f1 + . . . + Xn fn .
∑
For each 1 ≤ i ≤ n, let Xi = j xi,j with deg xi,j = j be the canonical decomposition of Xi into a sum of homogeneous elements: i.e., deg xi,j = j . Then
(13) f= ∞n
∑∑ xi,d−deg fi fi . d=0 i=1 Since f is homogeneous of degree deg(f ), only the d = deg(f ) in the right hand
side of (13) is nonzero, so
n
∑
f=
xi,deg f −deg fi fi ,
i=1 qed.
8.11.2. The Krull Intersection Theorem.
Theorem 221. Let R be a Noetherian ring, and I an ideal of R. Suppose there is
∩∞
an element x of R such that x ∈ n=1 I n . Then x ∈ xI .
Proof. The following miraculously short and simple proof is due to H. Perdry [Pe04].
Suppose I = ⟨a1 , . . . , ar ⟩. For each n ≥ 1, since x ∈ I n there is a homogeneous
degree n polynomial Pn (t1 , . . . , tr ) ∈ R[t1 , . . . , tr ] such that
x = Pn (a1 , . . . , an ).
By the Hilbert Basis Theorem (Theorem 218), R[t1 , . . . , tr ] is Noetherian. Therefore, deﬁnining Jn = ⟨P1 , . . . , Pn ⟩, there exists N such that JN = JN +1 . By Lemma
220 we may write
PN +1 = QN P1 + . . . + Q1 PN ,
with Qi homogeneous of degree i > 0. Plugging in ti = ai for 1 ≤ i ≤ n, we get
x = PN +1 (a1 , . . . , an ) = x (Q1 (a1 , . . . , an ) + . . . + QN (a1 , . . . , aN )) .
Since each Qi is homogeneous of positive degree, we have Qi (a1 , . . . , an ) ∈ I , qed! COMMUTATIVE ALGEBRA 143 Corollary 222. Let R be a Noetherian ring and I an ideal of R. Suppose that
either of the following holds:
(i) R is a domain and I is a proper ideal.
(ii) I ∩ contained in the Jacobson radical J (R) of R.
is
∞
Then n=1 I n = 0.
∩∞
Proof. Either way, let x ∈ n=1 I n and apply Theorem 221 to obtain an element
a ∈ I such that x = xa. Thus (a − 1)x = 0. Under assumption (i), we obtain either
a = 1 – so I = R, contradicting the properness of I – or x = 0. Under assumption
(ii), a ∈ J (R) implies a − 1 ∈ R× , so that we may multiply through by (a − 1)−1 ,
again getting x = 0.
E
a
∩xercise 8.30 (Su´rezAlvarez): Exhibit an ideal I in a Noetherian ring such that
∞
I n {0}. (Hint: idempotents!)
n=1
Exercise: Let R be the ring of all C ∞ functions f : R → R.
Let m = {f ∈ R  f (0) = 0}.
a) Show that m = xR is a maximal ideal of R.
b) Show that for all n ∈ Z+ , mn = {f ∈ R  f (0) = f ′ (0) = . . . = f (n) (0)}.
∩∞
c) Deduce that n=1 mn is the ideal of all smooth functions with identically zero
∩∞
Taylor series expansion at x = 0. Conclude that n=1 mn ̸= 0.
−1
d) Let f (x) = e x2 for x ̸= 0 and 0 for x = 0. Show that f ∈ f m.
/
e) Deduce that R is not Noetherian.
∪∞
1
Exercise: Let R = n=1 C[[t n ]] be the Puiseux series ring. Show that R is a
domain with a unique maximal ideal m and that for all n ∈ Z+ , mn = m. Deduce
from KIT that R is not Noetherian.
Remark: The preceding exercise will become much more routine when we study
valuation rings in §17. In that language, one can show that if ∩ is a valuation ring
R
∞
with divisible value group, then (R, m) is a local domain and n=1 mn = m.
8.12. Krull’s Principal Ideal Theorem.
Theorem 223. (The Principal Ideal Theorem, a.k.a. Krull’s Hauptidealsatz) Let
x be a nonunit in a Noetherian ring R, and let p be minimal among prime ideals
containing x. Then p has height at most one.
Remark: A prime p which is minimal among primes containing x will be called a
minimal prime over x. Note that an equivalent condition is that p is a minimal
prime in the quotient ring R/(x). Note also that if x is nilpotent, every prime of p
contains x so the height of any minimal prime is 0.
Our strategy of proof follows Kaplansky, who follows D. Rees. We need a preliminary result:
Lemma 224. Let u and y be nonzero elements in a domain R. Then:
a) The Rmodules ⟨u, y ⟩/(u) and ⟨u2 , uy ⟩/(u2 ) are isomorphic.
b) If we assume further that for all t ∈ R, tu2 ∈ (y ) implies tu ∈ (y ), then the
Rmodules (u)/(u2 ) and ⟨u2 , y ⟩/⟨u2 , uy ⟩ are isomorphic. 144 PETE L. CLARK Proof. a) The isomorphism is simply induced by multiplication by u.
b) The module (u)/(u2 ) is cyclic with annihilator (u), and conversely any such
module is isomorphic to R/(u). Moreover M := ⟨u2 , y ⟩/⟨u2 , uy ⟩ is also cyclic,
being generated simply by y . Certainly u annihilates M , so it suﬃces to show that
the annihilator is exactly (u). More concretely, given ky = au2 + buy , we must
deduce that k ∈ (u). But we certainly have au2 ∈ (y ), so by hypothesis au ∈ (y ),
say au = cy . Then ky = cuy + buy . Since 0 ̸= y in our domain R, we may cancel y
to get k = (c + b)u ∈ (u).
Proof of Krull’s Hauptidealsatz : Under the given hypotheses, assume for a contradiction that we have
p2 p1 p. Note ﬁrst that we can safely pass to the quotient R/p2 and thus assume that R is
a domain. Dually, it does not hurt any to localize at p. Therefore we may assume
that we have a Noetherian local domain R with maximal ideal m, an element x ∈ m,
and a nonzero prime ideal, say p, with x ∈ p
m, and our task is now to show
that this setup is impossible. Now for the clever part: let 0 ̸= y be any element of
p, and for k ∈ Z+ , let Ik denote the ideal of all elements t with txk ∈ (y ). Then
{Ik }∞ is an ascending chain of ideals in the Noetherian ring R so must stabilize,
k=1
say at k = n. In particular, tx2n ∈ (y ) implies txn ∈ (y ). Putting u = xn , we have
tu2 ∈ (y ) implies (tu) ∈ (y ).
Since m is a minimal prime over (x), the quotient ring T = R/(u2 ) has exactly
one prime ideal, m, and is therefore, by the AkizukiHopkins Theorem, an Artinian
ring, so that any ﬁnitely generated T module has ﬁnite length. In particular, M :=
⟨u, y ⟩/(u2 ), which can naturally be viewed as a T module, has ﬁnite length, and
hence so does its T submodule M ′ := ⟨u2 , y ⟩/(u2 ). Put N = ⟨u2 , y ⟩⟨u2 , uy ⟩. Then
ℓ(M ′ ) = ℓ(N ) + ℓ(⟨u2 , uy/(u2 )) = ℓ((u)/(u2 )) + ℓ(⟨u, y ⟩/(u)) = ℓ(M );
in the second equality we have used Lemma 224. The only way that M could have
the same length as its submodule M ′ is if ⟨u, y ⟩ = ⟨u2 , y ⟩, i.e., if there exist c, d ∈ R
such that u = cu2 + dy , or u(1 − cu) = −dy . Since u lies in the maximal ideal of
the local ring R, 1 − cu ∈ R× , and thus u ∈ (y ) ⊂ p. But m is minimal over x and
hence, being prime, also minimal over u = xn , contradiction!
Corollary 225. With hypotheses as in Theorem 223, suppose that x is not a zerodivisor. Then any prime p which is minimal over x has height one.
Exercise 8.31: Use the AkizukiHopkins theorem and Proposition 216 to give a
proof of Corollary 225.
As a second corollary of Theorem 223 we get the following striking structural result
about primes in a Noetherian ring. First a piece of notation: for any elements x, y
in a poset S we deﬁne the “interval” (x, y ) to be the set of all z ∈ S such that
x < z < y . For prime ideals p and q, by (p, q) we understand that the ambient
poset is the set of all prime ideals of R (which we are oﬃcially not yet allowed to
call Spec R....), i.e., it is the set of all prime ideals strictly between p and q.
Corollary 226. Let p ⊂ q be prime ideals in a Noetherian ring R. Then the
interval (p, q) is either empty or inﬁnite. COMMUTATIVE ALGEBRA 145 So if R is Noetherian with ﬁnitely many primes, its Krull dimension 0 or 1.
Again we need a small preliminary result.
Lemma 227. (Prime Avoidance) Let R be a ring, q an ideal of R and p1 , . . . , pn
∪
prime ideals. If q ⊂ i pi , then q ⊂ pi for some i.
Proof. We go by induction on n, the case n = 1 being trivial. Assume the result for
n − 1. Then if q were contained in the union of any n − 1 of the primes pi , then we
would be done by induction, so we may assume the existence of, for all 1 ≤ i ≤ n,
∪
an element xi ∈ pi ∩ (q \ j ̸=i pi ). If n = 2, put y = x1 + x2 ; then y is an element of
q but not of either p1 or p2 , a contradiction. In general, put y = x1 + x2 x3 · · · xn ;
then if y ∈ p1 then so is x2 · · · xn and since p1 is prime, for some i > 1, xi ∈ p1 ,
contradiction. Similarly if y ∈ pi for i > 1 then x1 ∈ pi , contradiction.
Exercise 8.32 (Kaplansky): Generalize the lemma to the case in which all but at
most two of the ideals pi are prime.
Proof of Corollary 226: As usual, by correspondence we may pass to R/p and therefore assume WLOG that p = 0. Suppose that for some n ≥ 1, [0, q] = {p1 , .∪. , pn }.
.
∪n
n
According to Lemma 227 we cannot then have q ⊂ i=1 pi , so choose x ∈ q \ i=1 pi .
Then q is a prime of R, of height at least 2, which is minimal over (x), contradicting
Theorem 223.
Theorem 228. (Generalized Principal Ideal Theorem) Let R be a Noetherian ring,
and let I = ⟨a1 , . . . , an ⟩ be a proper ideal of R. Suppose that I can be generated by
n elements. Let p be a minimal element of the set of all prime ideals containing I .
Then p has height at most n.
Proof. As usual, we may localize at p and suppose that R is local with p as its
maximal ideal. Suppose to the contrary that there exists a chain p = p0
p1
. . . pn+1 . Because R is Noetherian, we may arrange for (p1 , p) = ∅. Because p
is minimal over I , I cannot be contained in p1 ; without loss of generality we may
suppose that a1 is not in p1 . Put J := ⟨p1 , a1 ⟩; then J strictly contains p1 so p is
the unique prime of R containing J . So the ring R/J is an Artin local ring, and
then by Proposition 216 for suﬃciently large k we have pk ⊂ J . Then by taking t
to be suﬃciently large we can write, for 2 ≤ i ≤ n,
at = ci a1 + bi , ci ∈ R, bi ∈ p1 .
i
Put K = ⟨b2 , . . . , bn ⟩ ⊂ p1 . Since the height of p1 exceeds n − 1, by induction on n
we may assume that p1 properly contains a prime ideal Q which contains J . Now
the ideal Q′ := ⟨a1 , Q⟩ contains some power of each ai and therefore p is the unique
prime ideal containing Q′ . Therefore in the quotient domain R/Q, the prime p/Q is
minimal over the principal ideal Q′ /Q, so by Krull’s Hauptidealsatz p/Q has height
1. But on the other hand we have p/Q p1 /Q 0, a contradiction.
8.13. The Dimension Theorem.
The following is a very basic theorem about Noetherian rings:
Theorem 229. (Dimension Theorem) Let R be a Noetherian ring, and R[t] be the
univariate polynomial ring over R. Then
dim R[t] = dim R + 1. 146 PETE L. CLARK Remark: For a nonNoetherian ring R, one has the inequalities
dim R + 1 ≤ dim R[t] ≤ 2 dim R[t] + 1,
and indeed there are rings for which dim R[t] = 2 dim R + 1.
Of course, an immediate induction argument gives:
Corollary 230. Let k be a ﬁeld. Then
dim k [t1 , . . . , tn ] = n.
At this time we do not aspire to give a proof of Theorem 229. Neither will we
use it formally in the sequel, i.e., the proof of no future result will depend on it.
Nevertheless it is such a simple and natural fact that to deprive the reader of it
would be a serious blow against developing any intuition for the notion of dimension
of a Noetherian ring.
8.14. The ArtinTate Lemma.
Theorem 231. (ArtinTate [AT51]) Let R ⊂ T ⊂ S be a tower of rings such that:
(i) R is Noetherian,
(ii) S is ﬁnitely generated as an Ralgebra, and
(iii) S is ﬁnitely generated as a T module.
Then T is ﬁnitely generated as an Ralgebra.
Proof. Let x1 , . . . , xn be a set of generators for S as an Ralgebra, and let ω1 , . . . , ωm
be a set of generators for S as a T module. For all 1 ≤ i ≤ n, we may write
∑
(14)
xi =
aij ωj , aij ∈ T.
j Similarly, for all 1 ≤ i, j ≤ m, we may write
∑
bijk ωk , bijk ∈ T.
(15)
ωi ωj =
i,j,k Let T0 be the Rsubalgebra of T generated by the aij and bijk . Since T0 is a ﬁnitely
generated algebra over the Noetherian ring R, it is itself a Noetherian ring by the
Hilbert Basis Theoerem.
Now each element of S may be expressed as a polynomial in the xi ’s with Rcoeﬃcients. Making substitutions using (14) and then (15), we see S is generated
as a T0 module by ω1 , . . . , ωm , and in particular that S is a ﬁnitely generated T0 module. Since T0 is Noetherian, the submodule T is also ﬁnitely generated as a
T0 module. This immediately implies that T is ﬁnitely generated as a T0 algebra
and then in turn that T is ﬁnitely generated as an Ralgebra, qed!
9. Boolean rings
9.1. Deﬁnition and ﬁrst properties of Boolean rings.
Let R be a ring, not necessarily commutative, but with a multiplicative identity 1,
with the property that x2 = x for all x ∈ R. Then (1 + 1) = (1 + 1)2 = 1 + 1 + 1 + 1,
so 1 + 1 = 0. It follows that −x = x for all x ∈ R. Moreover for any x, y ∈ R,
(x + y ) = (x + y )2 = x2 + xy + yx + y 2 = x + y + xy + yx, so xy + yx = 0 or xy = yx.
Therefore such a ring is necessarily commutative. COMMUTATIVE ALGEBRA 147 With this remark in mind, deﬁne a Boolean ring to be a commutative ring with
identity such that x2 = x for all elements x.
Exercise 9.1: Show that the group of units of a Boolean ring is trivial.
Exercise 9.2: a) Show that any quotient ring of a Boolean ring is Boolean.
b) Show that any subring of a Boolean ring is Boolean.
Exercise 9.3: Show that a Boolean ring is absolutely ﬂat.
9.2. Boolean Algebras.
A Boolean ring is an object of commutative algebra. It turns out that there is a completely equivalent class of structures of an ordertheoretic nature, called Boolean
algebras. In some ways the concept of a Boolean algebra is more intuitive and
transparent – e.g., starting directly from the deﬁnition, it is perhaps easier to give
examples of Boolean algebras. Moreover it is not at all diﬃcult to see how to pass
from a Boolean ring to a Boolean algebra and conversely.
A Boolean algebra is a certain very nice partially ordered set (B, ≤). Recall
that for any partially orderet set B and any subset S , we have the notion of the
supremum sup S and the inﬁmum inf S . To deﬁne these it is convenient to extend
the inequality notation as follows: if S, T are subsets of B , we write
S<T
to mean that for all s ∈ S and t ∈ T , s < t, and similarly
S≤T
to mean that for all s ∈ S and t ∈ T , s ≤ t.
Then we say that z = sup S if S ≤ z and if w is any element of B with S ≤ w,
then z ≤ w. Similarly z = inf S if z ≤ S and if w is any element of B with w ≤ S
then w ≤ Z . For a given subset S , neither sup S nor inf S need exist, but if either
exists it it is plainly unique. In particular if sup ∅ exists, it is necessarily a bottom
element, called 0, and if inf ∅ exists, it is necessarily a top element called 0.
A partially ordered set (L, ≤) is called a lattice if for all x, y ∈ L, sup{x, y } and
inf {x, y } both exist. We give new notation for this: we write
x ∨ y := sup{x, y },
the join of x and y and
x ∧ y := inf {x, y },
the meet of x and y .
A lattice is said to be bounded if it contains a bottom element 0 and a top element
1: equivalently, sup S and inf S exist for every ﬁnite subset S .
Exercise 9.4: Let L be a lattice containing 0 and 1, and let x ∈ L. Then:
a) x ∨ 1 = 1,
b) x ∧ 1 = x,
c) x ∨ 0 = x,
d) x ∧ 0 = 0. 148 PETE L. CLARK A lattice L is complemented if it has a bottom element 0, a top element 1,
and for each x ∈ L there exists y ∈ L such that x ∨ y = 1, x ∧ y = 0.
A lattice is distributive if ∀x, y, z ∈ L,
(x ∨ y ) ∧ z = (x ∧ z ) ∨ (y ∧ z ),
(x ∧ y ) ∨ z = (x ∨ z ) ∧ (y ∨ z ).
Proposition 232. Let L be a distributive complemented lattice. Then for all x ∈ L,
the complement of x is unique.
Proof. Suppose that y1 and y2 are both complements to x, so
x ∨ y1 = x ∨ y2 = 1, x ∧ y1 = x ∧ y2 = 0.
Then
y2 = 1 ∧ y2 = (x ∨ y1 ) ∧ y2 = (x ∧ y2 ) ∨ (y1 ∧ y2 ) = 0 ∨ (y1 ∧ y2 ) = y1 ∧ y2 ,
so y2 ≤ y1 . Reasoning similarly, we get y1 ≤ y2 , so y1 = y2 .
By virtue of Proposition 232 we denote the complement of an element x in a distributive complemented lattice as x∗ .
Exercise 9.5: Show that for every element x of a distributive complemented lattice
we have (x∗ )∗ = x.
A Boolean algebra is a complemented distributive lattice with 0 ̸= 1.
Exercise 9.6: Show that DeMorgan’s Laws hold in any Boolean algebra B : for
all x, y ∈ B , we have
a) (x ∧ y )∗ = x∗ ∨ y ∗ and
b) (x ∨ y )∗ = x∗ ∧ y ∗ .
The shining example of a Boolean algebra is the powerset algebra 2S for a nonempty
set S . In the special case in which S  = 1, we denote the corresponding Boolean
algebra (the unique totally ordered set on two elements) simply as 2.
Not every Boolean algebra is isomorphic to a power set Boolean algebra.
Example: Let S be a set, and let Z (S ) ⊂ 2S be the collection of all ﬁnite and
coﬁnite subsets of S . It is easily checked that (Z (S ), ⊂) ⊂ (2S , ⊂) is a subBoolean
algebra. However, Z (S ) = S , so if S  = ℵ0 , then Z (S ) is not isomorphic to any
power set Boolean algebra.
Boolean algebras form a full subcategory of the category of partially ordered sets.
In other words, we deﬁne a morphism f : B → B ′ of Boolean algebras simply to be
an isotone (or orderpreserving) map: ∀x, y ∈ B, x ≤ y =⇒ f (x) ≤ f (y ). One can
(and sometimes does, e.g. for modeltheoretic purposes) also axiomatize Boolean
algebras as a structure (B, ∨, ∧, ∗, 0, 1), the point being that x ≤ y iﬀ x ∨ y = y iﬀ
x ∧ y = x, so the partial ordering can be recovered from the wedge or the join. COMMUTATIVE ALGEBRA 149 Proposition 233. The category of Boolean rings is equivalent to the category of
Boolean algebras.
In other words, we can deﬁne a functor F from Boolean rings to Boolean algebras
and a functor G from Boolean algebras to Boolean rings such that for every Boolean
ring R, R is naturally isomorphic to G(F (R)) and for every Boolean algebra B , B
is naturally isomorphic to F (G(B )).
Let us sketch the basic construction, leaving the details to the reader. Suppose
ﬁrst that R is a Boolean ring. Then we associate a Boolean algebra F (R) with the
same underlying set as R, endowed with the following operations: ∀x, y ∈ R,
x ∧ y = xy,
x∗ = 1 − x.
We should also of course deﬁne the join operation, but the point is that it is forced
on us be DeMorgan’s Laws:
x ∨ y = (x∗ ∧ y ∗ )∗ = x + y − xy.
Exercise 9.7: Check that (F (R), ∧, ∨, ∗) is indeed a Boolean algebra, and that the
bottom element 0 in F (R) (resp. the top element 1) is indeed the additive identity
0 (resp. the multiplicative identity 1).
Conversely, suppose that we have a Boolean algebra (B, ∧, ∨, ∗). Then we deﬁne a
Boolean ring G(B ) on the same underlying set B , by taking
x + y := (x ∧ y ∗ ) ∨ (y ∧ x∗ )
xy := x ∧ y.
Note that the addition operation corresponds to the Boolean operation “exclusive
or” or, in more settheoretic language, symmetric diﬀerence x∆y .
Exercise 9.8: check that (G(B ), +, ·) is indeed a Boolean ring with additive identity
the bottom element 0 of B and multiplicative identity the top element 1 of B .
Exercise 9.9:
a) Let R be a Boolean ring. Show that the identity map 1R on R is an isomorphism
of Boolean rings R → G(F (R)).
b) Let B be a Boolean algebra. Show that the identity map 1B on B is an isomorphism of Boolean algebras B → G(F (B )).
Exercise 9.10: Let X be a nonempty set, let BX be the Boolean algebra of subsets
of X , partially ordered by inclusion. Show that the corresponding Boolean ring
may be identiﬁed with the ring 2X of all functions from X to F2 under pointwise
addition and multiplication.
Exercise 9.11: Show that every ﬁnite Boolean algebra is isomorphic to a powerset algebra. Conclude that every ﬁnite Boolean ring R is isomorphic to the ring
of binary functions on a ﬁnite set of cardinality log2 (#R). In fact, try to show
this both on the Boolean algebra side and on the Boolean ring side. (Hint for
the Boolean ring side: use the decomposition into a direct product aﬀorded by an 150 PETE L. CLARK idempotent element.)
Exercise 9.12: Show that an arbitrary direct product of Boolean algebras (or, equivalently, Boolean rings) is a Boolean algebra (or...).
As we saw above, cardinality considerations already show that not every Boolean
algebra is the power set Boolean algebra, and hence not every Boolean ring is the
full ring of binary functions on some set X . However, in view of results like Cayley’s
theorem in basic group theory, it is a reasonable guess that every Boolean algebra is
an algebra of sets, i.e., is a subBoolean algebra of a power set algebra. We proceed
to prove this important result on the Boolean ring side.
9.3. Ideal Theory in Boolean Rings.
Proposition 234. Let R be a Boolean ring.
a) For all x ∈ N and all n ≥ 2, xn = x.
b) A Boolean ring is reduced, i.e., has no nonzeronilpotent elements.
c) Every ideal in a Boolean ring is a radical ideal.
Proof. a) The case n = 2 is the deﬁnition of a Boolean ring, so we may assume
n ≥ 3. Assume the result holds for all x ∈ R and all 2 ≤ k < n. Then xn =
xn−1 x = x · x = x.
b) If x ∈ R is such that xn = 0 for some positive integer n, then either n = 1 or
n ≥ 2 and xn = x; either way x = 0.
c) Let I be an ideal in the Boolean ring R. Then I = rad(I ) iﬀ R/I is reduced, but
R/I is again a Boolean ring and part b) applies.
The ring Z/2Z is of course a Boolean ring. It is also a ﬁeld, hence certainly a local
ring and an integral domain. We will shortly see that it is unique among Boolean
rings in possessing either of the latter two properties.
Proposition 235. a) The only Boolean domain is Z/2Z.
b) Every prime ideal in a Boolean ring is maximal.
Proof. a) Let R be a Boolean domain, and let x be an element of R. Then x(x − 1) =
0, so in a domain R this implies x = 0 or x = 1, so that R ∼ Z/2Z.
=
b) If p is a prime ideal in the Boolean ring R, then R/p is a Boolean domain,
hence – by part a) – is simply Z/2Z. But this ring is a ﬁeld, so p is maximal.
Proposition 236. Let R be a local Boolean ring. Then R ∼ Z/2Z.
=
Proof. Let m be the unique maximal ideal of R. By Proposition 235, m is moreover
the unique prime ideal of R. It follows from Proposition 121d) that m = nil R is
the set of all nilpotent elements, so by Proposition 234b) m = 0. Thus R is a ﬁeld
and thus, by Proposition 235 must be Z/2Z.
Exercise 9.13: Let R be a Boolean ring, let I be an ideal of R.
a) Show that x, y ∈ I =⇒ x ∨ y ∈ I .
b) Show in fact that ⟨x, y ⟩ = ⟨x ∨ y ⟩.
c) Deduce that any ﬁnitely generated ideal of R is principal.
Thus, for a Boolean ring R, all ideals of R are principal iﬀ R is Noetherian. But
in fact very few Boolean rings are Noetherian: we have already seen them all in
Exercise X.X. COMMUTATIVE ALGEBRA 151 Proposition 237. For a Boolean ring R, the following conditions are equivalent:
(i) R is ﬁnite.
(ii) R is Noetherian.
(iii) R has ﬁnitely many maximal ideals.
If these equivalent conditions hold, then R ∼ (Z/2Z)n , with n = log2 #R.
=
Proof. That (i) =⇒ (ii) is clear.
(ii) =⇒ (iii): Since R is Noetherian and prime ideals are maximal, by the AkizukiHopkins Theorem (Theorem 215) R is Artinian. Thus by Theorem 217 ⊕ is a ﬁnite
R
n
product of local Boolean rings, and thus ﬁnally by Proposition 236 R ∼ i=1 Z/2Z.
=
(iii) =⇒ (i): Suppose that R has precisely N < ∞ maximal ideals and suppose for a
contradiction that R is inﬁnite. Then we may choose x ∈ R \{0, 1}, i.e., a nontrivial
idempotent. This leads to a direct product decomposition R = xR × (1 − x)R. Here
xR and (1 − x)R are subrings of R, hence at least one of them is an inﬁnite Boolean
ring. It follows that this decomposition process can be continued indeﬁnitely, or
⊕N +1
more precisely until we get to the point of writing R = i=1 Ri as a product of
N + 1 Boolean rings. For i = 1, . . . , N + 1, let mi be a maximal ideal of Ri . Then
∏
for i = 1, . . . , N + 1, Mi = j ̸=i Rj × mi are distinct maximal ideals of R, which
thus has at least N + 1 maximal ideals, contradiction.
Lemma 238. Let m be an ideal in the Boolean ring R. TFAE:
(i) m is maximal.
(ii) For all x ∈ R, either x ∈ R or 1 − x ∈ R (and not both!).
Proof. (i) =⇒ (ii): Of course no proper ideal in any ring can contain both x
and 1 − x for then it would contain 1. To see that at least one must lie in m, it is
certainly no loss to assume that x is neither 0 nor 1, hence R = xR × (1 − x)R.
Recall that in a product R1 × R2 of rings, every ideal I is itself a product I1 × I2 ,
where Ii is an ideal of Ri . Then R/I ∼ R1 /I1 × R2 /I2 . So I is prime iﬀ R/I is a
=
domain iﬀ either (i) I1 is prime in R1 and I2 = R2 or (ii) I1 = R1 and I2 is prime in
R2 . In particular, every maximal ideal of R1 × R2 contains either R1 or R2 : done.
(i) =⇒ (ii):39 We prove the contrapositive: if m is not maximal, there exists a
maximal ideal M properly containing m. Let x ∈ M \ m. Since M is proper, it does
not also contain 1 − x, hence neither does the smaller ideal m.
Exercise 9.14 (Kernel of a homomorphism): Let f : R → F2 be a homomorphism
of Boolean rings.
a) Show that Ker f is ∨closed: if x, y ∈ Ker f , then x ∨ y ∈ Ker f .
b) Show that Ker f is downwardclosed: if x ∈ Ker f and y ≤ x, then y ∈ Ker f .
c) Explain why parts a) and b) are equivalent to showing that Ker f is an ideal of
the Boolean ring R.
d) Show that Ker f is in fact a maximal ideal of R.
e) Conversely, for every maximal ideal m of R, show that R/m = F2 and thus the
quotient map q : R → R/m is a homomorphism from R to F2 .
Exercise 9.15 (Shell of a homomorphism): Let f : R → F2 be a homomorphism of
Boolean rings. Deﬁne the shell Sh f to be f −1 (1).
a) Show that Sh f is wedgeclosed: if x, y ∈ Sh f , so is x ∧ y .
b) Show that Sh f is upwardclosed: if x ∈ Sh f and x ≤ y , then y ∈ Sh f .
39Note that this direction holds in any ring. 152 PETE L. CLARK c) A nonempty, proper subset of a Boolean algebra which is wedgeclosed and
upwardclosed is called a ﬁlter, so by parts a) and b) Sh f is a ﬁlter on B . Show
that in fact it is an ultraﬁlter on B , i.e., that it is not properly contained in any
other ﬁlter. (Suggestion: use Lemma 238.)
d) Show that every ultraﬁlter on B is the shell of a unique homomorphism of
Boolean algebras f : B → F2 .
9.4. The Stone Representation Theorem.
Let R be a Boolean ring. We would like to ﬁnd an embedding of R into a Boolean
ring of the form 2X . The key point of course, is to conjure up a suitable set X .
Can we ﬁnd any clues in our prior work on Boolean rings?
indent Well, ﬁnite Boolean rings we understand: the proof of Proposition 237 gives
⊕
us that every ﬁnite Boolean ring is of the form
i∈X Z/2Z, where the elements
⊕
of X correspond to the maximal ideals of R. The isomorphism i∈X Z/2Z ∼ 2X
=
amounts to taking each element x of R and recording which of the maximal ideals
it lies in: namely, x lies in the ith maximal ideal mi of all elements having a zero
in the ith coordinate iﬀ its image in the quotient R/mi = Z/2Z is equal to 0.
This motivates the following construction. For any Boolean ring, let M (R) be
the set of all maximal ideals of R, and deﬁne a map E : R → M (R) by letting E (x)
be the set of maximal ideals of R which do not contain x. This turns out to be
very fruitful:
Theorem 239. (Stone Representation Theorem) Let R be a Boolean ring and
M (R) the set of maximal ideals. The map E : R → 2M (R) which sends an element
x of R to the collection of all maximal ideals of R which do not contain x is an
injective homomorphism of Boolean rings. Therefore R is isomorphic to the Boolean
ring associated to the algebra of sets E (R) ⊂ 2M (R) .
In particular this shows that every Boolean algebra is an algebra of sets.
Proof. Step 1: We check that the map E is a homomorphism of Boolean algebras.
Above we saw E (0) = ∅; also E (1) = M (R). Also, for x, y ∈ R, E (xy ) is the set of
maximal ideals which do not contain xy ; since maximal ideals are prime this is the
set of maximal ideals which contain neither x nor y , i.e., E (x) ∩ E (y ) = E (x) · E (y ).
Finally, E (x) + E (y ) = E (x) ∆ E (y ) is the set of maximal ideals which contain
exactly one of x and y , whereas E (x + y ) is the set of maximal ideals not containing
x + y . For m ∈ M (R), consider the following cases:
(i) x, y ∈ m. Then m is not in E (x) ∆ E (y ). On the other hand x + y ∈ m, so m is
not in E (x + y ).
(ii) Neither x nor y is in m. Certainly then m is not in E (x) ∆ E (y ). On the other
hand, remembering that R/m ∼ Z/2Z, both x and y map to 1 in the quotient, so
=
x + y maps to 1 + 1 = 0, i.e., x + y ∈ m, so m is not in E (x + y ).
(iii) Exactly one of x and y lies in m. Then m ∈ E (x) ∆E (y ) and as above, x + y
maps to 1 in R/m, so x + y is not in m and m ∈ E (x + y ).
Step 2: We show that the map E is injective. In other words, suppose we have two
elements x and y of R such that a maximal ideal m of R contains x iﬀ it contains COMMUTATIVE ALGEBRA y . Then
(x) = rad(x) = ∩
m∈M (R)  x∈ m m= ∩ 153 m = rad(y ) = (y ). m∈M (R)  y ∈ m So there exist a, b ∈ R with y = ax, x = by , and then
x = by = by 2 = xy = ax2 = ax = y. Let us comment a bit on the proof. Although Step 1 is longer, it is clearly rather
routine. They key is of course that E gives an embedding, which as we saw is equivalent to the much gutsier statement that an element of a Boolean ring is entirely
determined by the family of maximal ideals containing it. From the standpoint
of the more “conventional” rings one encounters in number theory and algebraic
geometry, this is a very strange phenomenon. First of all, it can only be true in
a ring R which has trivial unit group, since if u is a nontrivial unit of course we
will not be able to distinguish 1 and u using ideals! Moreover it implies that every prinicpal ideal is radical, which is impossible in any integral domain. Finally
it implies that every radical ideal is the intersection of the maximal ideals which
contain it, a property which we will meet later on the course: such rings are called
Jacobson rings.
9.5. Boolean spaces.
For reasons which I hope will shortly be made clear, we will now digress a bit to
talk (not for the ﬁrst or last time!) about topological spaces. Following Bourbaki,
for us compact means quasicompact and Hausdorﬀ. Further a locally compact
space is a Hausdorﬀ space in which each point admits a local base of compact
neighborhoods. A subset of a topological space is clopen if it is both closed and
open.
Recall that a topological space X is totally disconnected if the maximal connected subsets of X are the singleton sets {x}. Note that a totally disconnected
space is necessarily separated (older terminology that I am not fond of: T1 ): i.e.,
singleton sets are closed. Indeed, the closure of every connected set is connected,
so the closure of a nonclosed point would give a connected set which is larger than
a point. On the other hand a space X is zerodimensional if it admits a base of
clopen sets.
Proposition 240. Let X be a locally compact space. Then X is totally disconnected
iﬀ it is zerodimensional.
Proof. Exercise! (This is not used in the sequel.)
A space X is called Boolean40 if it is compact and zerodimensional; in particular
a Boolean space admits a base for the topology consisting of compact open sets.
Exercise 9.16: a) A ﬁnite space is Boolean iﬀ it is discrete.
b) A Boolean space is discrete iﬀ it is ﬁnite.
c) An arbitrary direct product of Boolean spaces is Boolean.
d) The usual Cantor space is homeomorphic to a countably inﬁnite direct product
40There are many synonyms: e.g. Stone space, proﬁnite space. 154 PETE L. CLARK of copes of a discrete, twopoint space and thus is a Boolean space.
Exercise 9.17: Show that a topological space is Boolean iﬀ it is homeomorphic
to an inverse limit of ﬁnite, discrete spaces.
To every topological space X we may associate a Boolean algebra: namely, the
subalgebra of 2X consisting of compact open subsets. Thus in particular we may
associate a Boolean ring, say C (X ), the characteristic ring of X . (We also ourselves to pass between C (X ) and the associated Boolean algebra on the same set
and call the latter the characteristic algebra.)
Exercise 9.18: Show that the assignment X → C extends to a contravariant functor
from the category of topological spaces to the category of Boolean rings. (In other
words, show that a continuous map f : X → Y of topological spaces induces a
“pullback” homomorphism C (f ) : C (Y ) → C (x) of Boolean rings.)
If X is itself a Boolean space, then the characteristic algebra C (X ) is indeed characteristic of X in the following sense.
Proposition 241. Let X be a Boolean space, and let A be a Boolean algebra of
subsets of X which is also a base for the topology of X . Then A = C (X ).
Proof. By hypothesis the elements of A are open sets in X . Moreover, since A is
closed under complementation, the elements are also closed. Thus A ⊂ C (X ).
Conversely, suppose Y ∈ C (X ). Since Y is open and A is a base for the topology
on X , for each y ∈ Y there is Ay ∈ A with y ∈ Ay ⊂ Y . Thus {Ay }y∈Y is an
open cover for Y . But Y is also closed in a ∪
compact space hence itself compact, so
n
we may extract a ﬁnite subcover, say Y = i=1 Ayi . Since A is a subalgebra, it is
closed under ﬁnite unions, so Y ∈ A. Thus C (X ) ⊂ A.
To every Boolean ring R we may associate a Boolean space: namely there is a
natural topology on M (R), the set of maximal ideals of R, with respect to which
M (R) is a Boolean space. This topology can be described in many ways.
First Approach: by the Stone Representation Theorem we have an embedding
R → 2M (R) and thus every element x ∈ R determines a function x : M (R) → F2 .
We may endow F2 with the discrete topology (what else?) and then give M (R) the
initial topology for the family of maps {x : M (R) → F2 }x∈R , that is the ﬁnest
topology which makes each of these maps continuous.
Here is a more concrete description of this initial topology: for each x ∈ R, put
Ux = {m ∈ M (R)  x ∈ m}
/
Vx = {m ∈ M (R)  x ∈ m}.
Then the topology in question is the one generated by {Ux , Vx }x∈X .
Second Approach: for any Boolean ring R, to give a maximal ideal m of R is
equivalent to giving a homomorphism of Boolean rings f : R → F2 . Namely, to a
maximal ideal m we associate the quotient map, and to f : R → F2 we associate the
kernel f −1 (0). In this way we get an embedding ι : M (R) → 2R . Now we endow COMMUTATIVE ALGEBRA 155 each copy of F2 with the discrete topology and 2R with the product topology: this
makes it into a Boolean space.
Lemma 242. The image ι(M (R)) of ι is a closed subspace of 2R .
Exercise 9.19: Prove Lemma 242.
Thus if we endow M (R) with the topology it inherits via the embedding ι, it
is itself a Boolean space.
Exercise 9.20: Show that the topology on M (R) deﬁned via Lemma 242 coincides with the initial topology on M (R) deﬁned above.
It turns out to be important to consider a distinguished base for the topology
on M (R), which we now deﬁne. Since for a prime ideal m we have xy ∈ m ⇐⇒
/
x ∈ m andy ∈ m, we have for all x, y ∈ R that
/
/
Ux ∩ Uy = Uxy .
Moreover, by Lemma 238, M (R) \ Vx = Ux . It follows that the {Ux }x∈X form a
base of clopen sets for the topology on M (R).
To show the utility of this base, let us use it to show directly that M (R) is a
Boolean space.
Hausdorﬀ: Let m1 and m2 be distinct maximal ideals of R. Choose x ∈ m2 \ m1 , so
by Lemma 238 1 − x ∈ m1 \ m2 . Thus m1 ∈ Ux m2 ∈ U1−x and
Ux ∩ U1−x = Ux(1−x) = U0 = ∅,
so we have separated m1 and m2 by open sets.
Quasicompact: As is wellknown, it is enough to check quasicompactness of a
space using covers by elements of any ﬁxed base. We certainly have a preferred
base here, namely {Ux }x∈X , so let’s use it: suppose that we have a collection
∪
{xi }i∈I such that i∈I Uxi = M (R). Now again (and not for the last...) we exploit
the power of DeMorgan:
∪
∪
∩
M (R ) =
Uxi = (M (R) \ Vxi ) = M (R) \
V xi ,
∩ i i i so that i Vxi = ∅. This means that there is no maximal ideal containing every
xi . But that means that the ideal generated by the xi ’s contains 1:∩there exists a
∑
ﬁnite subset J ⊂ I and aj ∈ R such that j aj xj = 1, and thus j ∈J Vxj = ∅:
∪
equivalently j ∈J Uxj = M (R).
Thus we have shown that the correspondence R → M (R) associates to every
Boolean ring a Boolean topological space, its Stone space.
Exercise 9.21: Show that the assignment R → M (R) extends to a functor from
the category of Boolean rings to the category of Boolean spaces. 156 PETE L. CLARK 9.6. Stone Duality.
Theorem 243. (Stone Duality): The functors C and M give a duality between the
category of Boolean spaces and the category of Boolean algebras. More concretely:
a) For every Boolean algebra B , the map B → C (M (B )) given by x ∈ B → Ux is
an isomorphism of Boolean algebras.
b) For every Boolean space X , the map m : X → M (C (X )) given by x ∈ X →
mx := {U ∈ C (X )  x ∈ U } is a homeomorphism of Boolean spaces.
/
Proof. a) The map e : x ∈ B → Ux ∈ 2M (B ) is nothing else than the embedding e
of the Stone Representation Theorem. In particular it is an embedding of Boolean
algebras. Its image e(B ) is a subalgebra of the characteristic algebra of the Boolean
space M (B ) which is, by deﬁnition, a base for the topology of M (B ). By Proposition 241 we have e(B ) = C (M (B )) so e is an isomorphism of Boolean algebras.
b) First we need to show that mx is a maximal ideal in the characteristic ring C (X ).
It seems more natural to show this on the Boolean algebra side, i.e., to show that mx
is downward closed and unionclosed. Indeed, U ∈ mx means x ∈ U , so if V ⊂ X
/
then certainly x ∈ V , i.e., V ∈ mx ; moreover, U, V ∈ mx ⇐⇒ x ∈ U and x ∈
/
/
/
V ⇐⇒ x ∈ U ∪ V ⇐⇒ U ∪ V ∈ mx . Thus mx is an ideal of C (X ). Applying
/
Lemma 238, one easily sees that is maximal, so the map m is welldeﬁned.
indent The injectivity of m follows immediately from the Hausdorﬀ property of X .
Surjectivity: Let m ∈ M (C (X )). By Exercise X.X, we may identify m with a
−
homomorphism of Boolean algebras fm : C (X ) → F2 . Let F = fm 1 (1) be the
shell of fm , an ultraﬁlter on the Boolean algebra of sets C (X ). In particular F is
wedgeclosed, i.e., it is a family of clopen subsets of the compact space X satisfy∩
ing the ﬁnite intersection property. Therefore there exists x ∈ U ∈F U . On the
other hand, the collection Fx of all clopen sets in X containing x is also a ﬁlter on
C (X ) with F ⊂ Fx . But since F is an ultra ﬁlter – i.e., a maximal ﬁlter – we have
F = Fx . Thus m and Fx are respectively the kernel and shell of the homomorphism
f : C (X ) → F2 , so
m = C (X ) \ Fx = {U ∈ C (X )  x ∈ U } = m(x).
/
Finally, since m is surjective, we have that for each A ∈ C (X ),
{U ∈ M (C (X ))  A ∈ U } = {m(x)  x ∈ A},
so that m maps the base C (X ) for the topology on X onto the base C (M (C (X ))).
Exercise 9.22: Let X be a topological space, and let C (X, 2) be the ring of all
continuous functions f : X → F2 (F2 being given the discrete topology).
a) Show that C (X, 2) is a Boolean ring.
b) Suppose that X = M (R) is the maximal ideal space of the Boolean ring R.
Show that C (X, 2) is canonically isomorphic to R itself. Thus every Boolean ring
is the ring of continuous Booleanvalued functions on its Stone space of maximal
ideals.
9.7. Topology of Boolean Rings.
Proposition 244. Let R be a Boolean ring and m a maximal ideal of R. TFAE:
(i) m is an isolated point in the Stone space M (R).
(ii) m = Rx is a principal ideal. COMMUTATIVE ALGEBRA 157 Exercise 9.23: Prove Proposition 244.
A Boolean ring R is atomic if for every x ̸= 1 there exists a principal maximal ideal m with x ∈ m.
Exercise 9.24: For any nonempty set S , show that 2S = ∏
s ∈S Z/2Z is atomic. A Boolean ring is called atomless if it contains no maximal principal ideals.
Exercise 9.25: Show that a Boolean algebra B is atomless if for all x ∈ B , if
x < 1, there exists y ∈ B with x < y < 1.
Proposition 245. A Boolean ring R is atomless iﬀ its Stone space M (R) is perfect, i.e., without isolated points.,
Exercise 9.26: Prove Proposition 245.
Corollary 246. Any two countably inﬁnite atomless Boolean rings are isomorphic.
Exercise 9.27: Prove Corollary 246. (Suggestion: show that the Stone space of any
countably inﬁnite atomless Boolean ring is isomorphic to the Cantor set.)
Exercise 9.28 (for those who know some model theory):
a) Show that there is a ﬁrst order theory in the language (∨, ∧, ∗, 0, 1) whose models
are precisely the atomless Boolean algebras.
b) Use Vaught’s Test to show that this theory is complete.
Exercise 9.29: Let S be a nonempty set and consider the Boolean ring R = 2S =
∏
s∈S Z/2Z.
a) Show that there is a natural bijective correspondence between elements of S and
principal maximal ideals of R.
b) Deduce that there is an embedding ι : S → M (R) such that the induced topology on S is discrete.
c) Show that ι(S ) is dense in M (R). (Hint: R is atomic.)
d) Show that ι is a homeomorphism iﬀ S is ﬁnite.
e)* Show that ι is none other than the StoneCech compactiﬁcation of the
discrete space X .
10. Primary Decomposition
10.1. Some preliminaries on primary ideals.
Recall that a proper ideal q of a ring R is primary if for all x, y ∈ R, xy ∈ q
implies x ∈ q or y n ∈ q for some n ∈ Z+ .
Exercise 10.1: a) Show that a prime ideal is primary. (Trivial but important!)
b) Show that an ideal q of R is primary iﬀ every zerodivisor in R/q is nilpotent.
Neither the deﬁnition or primary ideal nor the characterization given in the above
exercise is particularly enlightening, so one natural question is: which ideals are
primary? (And, of course, another natural question is: what’s the signiﬁcance of 158 PETE L. CLARK a primary ideal?) Here are some simple results which give some information on
primary ideals, suﬃcient to determine all the primary ideals in some simple rings.
Proposition 247. Let q be an ideal in a ring R. If r(q) = m is a maximal ideal,
then q is primary. In particular, any power of a maximal ideal is primary.
Proof. Since r(q) is the intersection of all prime ideals containing q, if this intersection is a maximal ideal m, then m is the unique prime ideal containing q and
R/q is a local ring with nil(R/q) = J (R/q) = m/q. In such a ring an element is a
zerodivisor iﬀ it is a nonunit iﬀ it is nilpotent, so q is primary. The “in particular”
follows since by Proposition 122f), r(mn ) = r(m) = m.
Proposition 248. If q is a primary ideal, then its radical r(q) is a prime ideal,
the smallest prime ideal containing q.
Proof. Let xy ∈ r(q), so that (xy )m = xm y m ∈ p for some m ∈ Z+ . If xm is in q
then x ∈ r(q), so assume that xm is not in q. Then y m is a zero divisor in R/q,
so by deﬁnition of primary there exists n ∈ Z+ such that (y m )n ∈ q, and then
y ∈ r(q). The second statement holds for any ideal I whose radical is prime, since
r(I ) is the intersection of all prime ideals containing I .
A primary ideal is said to be pprimary if its radical is the prime ideal p.
∩n
Lemma 249. If q1 , . . . , qn are pprimary ideals, then q = i=1 qi is also pprimary.
Proof. Let x, y be elements of the ring R such that xy ∈ q and x ∈ R \ q. Then
∏n
for all 1 ≤ i ≤ n, there exists ai ∈ Z+ such that y ai ∈ Ii , and then y i=1 ai ∈ q, so
q is primary. Moreover, by Proposition 122b),
n
n
n
∩
∩
∩
r(q) = r( qi ) =
r(qi ) =
q = q.
i=1 i=1 i=1 Exercise 10.2: Give an example of primary ideals q, q′ such that q ∩ q′ is not primary.
Proposition 250. If q is a primary ideal, the quotient ring R/q is connected.
Proof. Indeed, a ring is disconnected if and only if it has an idempotent element e
diﬀerent from 0 or 1. Such an element is certainly not nilpotent en = e for all n –
but is a zerodivisor, since e(1 − e) = e − e2 = 0.
Exercise 10.3: Let k be a ﬁeld, let R = k [x, y ] and put I = (xy ). Show that I is
not primary but “nevertheless” R/I is connected.
Example: We will ﬁnd all primary ideals in the ring Z of integers. Evidently
(0) is prime and hence primary. If q is any nonzero primary ideal, then its radical
p = r(q) is a nonzero prime ideal, hence maximal. So, combining Propositions 247
and 248 we ﬁnd that a nonzero ideal in Z is primary iﬀ its radical is maximal.
Moreover, for any prime power (pn ), r((pn )) = r((p)) = (p) is maximal – we use
here the elementary and (we hope) familiar fact that if p is a prime number, (p) is
a prime ideal (Euclid’s Lemma); such matters will be studied in more generality in
§X.X on factorization – so (pn ) is a primary ideal. Conversely, if n is divisible by
more than one prime power, then applying the Chinese Remainder Theorem, we
get that Z/n is disconnected. COMMUTATIVE ALGEBRA 159 Exercise 10.4: a) Let R be an integral domain for which each nonzero ideal is
a (ﬁnite, of course) product of maximal ideals. Use the above argument to show
that an ideal q of R is primary iﬀ it is a prime power.
b) (For those who know something about PIDs) Deduce in particular that primary
= prime power in any principal ideal domain.
Remark: Consider the following property of an integral domain:
(DD) Every ideal can be expressed as a product of prime ideals.
This is a priori weaker than the hypothesis of Exericse X.Xa). Later we will devote
quite a lot of attention to the class of domains satisfying (DD), the Dedekind
domains. Among their many properties is that a Dedekind domain is (either a
ﬁeld or) a domain in which each nonzero prime ideal is maximal. Thus in fact the
hypothesis of Exercise 10.4a) is equivalent to assuming that R is a Dedekind domain.
Remark(ably): Another characterization theorem says that any Noetherian domain
in which each primary ideal is a prime power is a Dedekind domain. In particular,
any polynomial ring k [x1 , . . . , xn ] in 2 ≤ n < ∞ variables over a ﬁeld admits primary ideals which are not prime powers.
The following exercise gives an even simpler (and more explicit) example of a ring
R and a primary ideal q of R which is not a prime power.
√
Exercise 10.5: Let R = Z [t]/(t2 + 3) (or, equivalently, Z[ −3]). Let q = (2).
a) Show that there is a unique ideal p2 with R/p2 = Z/2Z. Evidently p2 is maximal.
b) Show that r(q) = p2 , and deduce that I is primary.
c) Show that q is not a prime power, and indeed, cannot be expressed as a product
of prime ideals.41
The ring R of Exercise 10.5 is a good one to keep in mind: it is simple enough to be
easy to calculate with, but it already displays some interesting general phenomena.
This is a Noetherian domain in which every nonzero prime ideal is maximal. It
is therefore “close” to being a Dedekind domain but it does not satisfy one other
property (“integral closure”) which will be studied later. It will turn out to be an
immediate consequence of the main result of this section that, notwithstanding the
fact that there are ideals which do not factor into a product of primes, nevertheless
√
every proper ideal in R = Z[ −3] can be written as a product of primary ideals.
This, ﬁnally, is some clue that the notion of a primary ideal is a fruitful concept.
Having seen examples of a primary ideals which are not prime powers, what about
the converse? Is it at any rate the case that any prime power is a primary ideal?
We know that this is indeed the case for powers of a maximal ideal. However, the
answer is again negative in general: 41Suggestion: show that R is a ring with ﬁnite quotients, and use properties of the norm
function I  = #R/I . 160 PETE L. CLARK Example (AtiyahMacDonald, p. 51): Let k be a ﬁeld; put R = k [x, y, z ]/(xy − z 2 ).
Denote by x, y , and z the images of x, y, z in R. Put p = ⟨x, z ⟩. Since R/p =
k [x, y, z ]/(x, z, xy − z 2 ) = k [y ] is a domain, p is a prime ideal. Now consider the
ideal p2 : we have xy = z 2 ∈ p2 , but x ∈ p2 and y ∈ p = r(p2 ), so p2 is not primary.
/
/
10.2. Primary decomposition, Lasker and Noether.
Let R be a ring and I an ideal of R. A primary decomposition of I is an
∩n
expression of I as a ﬁnite intersection of primary ideals, say I = i=1 qi .
An ideal which admits at least one primary decomposition is said to be decomposable. This is not a piece of terminology that we will use often, but the reader
should be aware of its existence.
For any ring R, let us either agree that R itself admits the “empty” primary decomposition or that R has no primary decomposition (i.e., it doesn’t matter either
way) and thereafter restrict our attention to proper ideals.
It may not be too surprising that not every ideal in every ring admits a primary
decomposition. Indeed, we will see later that if R is a ring for which (0) admits a
primary decomposition, then the ring R has ﬁnitely many minimal primes.
The ﬁrst important result in this area was proved by Emanuel Lasker in 1905,
roughly in the middle of his 27 year reign as world chess champion. Here it is.
Theorem 251. (Lasker [Las05]) Let R be a polynomial ring in ﬁnitely many variables over a ﬁeld. Then every proper ideal I of R admits a primary decomposition.
Lasker’s proof of this theorem was a long and intricate calculation. As we will
shortly see, a broader perspective yields considerably more for considerably less
eﬀort. In Lasker’s honor a ring R in which every proper ideal admits a primary
decomposition is called a Laskerian ring.
Exercise 10.6: If R is Laskerian and I is an ideal of R, then R/I is Laskerian.
Combining Lasker’s theorem with this Exercise, we get that every ﬁnitely generated
algebra over a ﬁeld admits a primary decomposition. This result is of fundamental
(indeed, foundational) importance in algebraic geometry.
However, in 1921 Lasker’s triumph was undeniably trumped by Emmy Noether.
Theorem 252. (Noether) Any proper ideal in a Noetherian ring admits a primary
decomposition.
In other words, a Noetherian ring is Laskerian. Therefore Lasker’s Theorem is
an immediate consequence of Noether’s Theorem together with the Hilbert Basis
Theorem, which we recall, was proved in 1888 and whose remarkably short and
simple – but nonconstructive – proof engendered ﬁrst controversy and later deep
admiration. The same is true for Noether’s theorem: it is from this theorem, and
the ridiculous simplicity of its proof, that Noetherian rings get their name. COMMUTATIVE ALGEBRA 161 Indeed, the result will practically prove itself once we introduce one further concept.
An ideal I is irreducible if whenever I is written as an intersection of two ideals
– i.e., I = J ∩ K – then I = J or I = K .
Exercise 10.7: Let I be a proper ideal in a principal ideal domain R. TFAE:
(i) I is primary.
(ii) I is irreducible.
(iii) I is a prime power: there exists a in R and n ∈ Z+ such that (a) is prime and
I = (a)n = (an ).
Dangerous bend: It follows therefore that if I = (x) is a principal ideal, then I
being irreducible as an ideal is not the same as the element x being irreducible,
even in very nice rings. For instance, it follows from Exercise X.X that an ideal
I = (n) in Z is irreducible iﬀ it is primary iﬀ n = pr is a prime power, whereas of
course n is irreducible as an element iﬀ n = p is prime. For this reason, we are not
thrilled with the term “irreducible.” We contemplated using the term “indecomposable” instead, but this ought to mean “not decomposable”, which is doesn’t.
At last we have decided not to try to “correct” the terminology, as we will not be
using it in the rest of these notes anyway.
Proposition 253. a) A prime ideal is irreducible.
b) An irreducible ideal in a Noetherian ring is primary.
Proof. a) Let p be a prime ideal, and write p = I ∩ J . Since then p ⊃ IJ , by
Proposition 118 we have p ⊃ I or p ⊃ J ; WLOG say p ⊃ I . Then p = I ∩ J ⊂ I ⊂ p,
so that we must have I = p.
b) By passage to the quotient, it is enough to assume that the 0 ideal is irreducible
and show that it is primary. So, suppose that xy = 0 and x ̸= 0. Consider the
chain of ideals
ann(y ) ⊂ ann(y 2 ) ⊂ . . . ⊂ ann(y n ) ⊂ . . . .
Since R is Noetherian, this chain stabilizes: there exists n such that ann(y n ) =
ann(y n+k ) for all k . We claim that (x) ∩ (y n ) = 0. Indeed, if a ∈ (x) then
ay = 0, and if a ∈ (y n ) then a = by n for some b ∈ R, hence by n+1 = ay = 0, so
b ∈ ann(y n+1 ) = ann(y n ), hence a = by n = 0. Since the (0) ideal is irreducible, we
must then have y n = 0, and this shows that (0) is primary.
Proof of Noether’s Theorem : Let I be a proper ideal in the Noetherian ring R.
We claim that I is a ﬁnite intersection of irreducible ideals; in view of Proposition
253 this gives the desired result. But the proof of this is a trivial application of
Noetherian induction: suppose that the set of proper ideals which cannot be written
as a ﬁnite intersection of irreducible ideals is nonempty, and choose a maximal
element I . Then I is reducible, so we may write I = J ∩ K where each of J
and K is strictly larger than I . But being strictly larger than I each of J and
K can be written as a ﬁnite intersection of irreducible ideals, and hence so can I .
Contradiction!
10.3. Irredundant primary decompositions.
If an ideal can be expressed as a product of prime ideals, that product is in fact 162 PETE L. CLARK unique. We would like to have similar results for primary decomposition. Unfortunately such a uniqueness result is clearly impossible. Indeed, if I = q1 ∩ . . . ∩ qn is
a primary decomposition of I and p is any prime containing I , then q1 ∩ . . . ∩ qn ∩ p
is also a primary decomposition, and clearly a diﬀerent one if p ̸= qi for any i. A
proper ideal I may well be contained in inﬁnitely many primes – e.g. by X.X this
occurs with I = (0) for any Noetherian domain of dimension at least 2 – so there
may well be inﬁnitely many diﬀerent primary decompositions.
But of course throwing in extra primes is both frivolous and wasteful. The following deﬁnition formalizes the idea of a primary decomposition which is “frugal”
in two reasonable ways.
A primary decomposition is said to be irredundant42 (or minimal, or reduced)
if both of the following properties hold:
(IPD1) For all i ̸= j , r(qi ) ̸= r(qj ). ∩
(IPD2) For all i, qi does not contain j ̸=i qj .
The following simple and important result tells us that if wastefulness succeeds,
then so does frugality.
Lemma 254. An ideal which admits a primary decompoistion admits an irredundant primary decomposition.
Proof. By Lemma 249, we may replace any collection of primary ideals qi with a
common radical with their intersection and still have a primary ideal, thus satisfying (IPD1). Then if (IPD2) is not satisﬁed, there is some qi which contains
the intersection of all the other qj ’s, hence it can simply be removed to obtain a
primary decomposition satisfying (IPD1) and with a smaller number of primary
ideals. Clearly proceeding in this way, we eventually arrive at an irredundant primary decomposition.
The question is now to what extent an irredundant primary decomposition is
unique. The situation here is signiﬁcantly better: although the primary decomposition is not in all cases unique, it turns out that there are some important
quantities which are deﬁned in terms of a primary decomposition and which can
be shown to be independent of the choice of irredundant decomposition, i.e., are
invariants of the ideal. Such uniqueness results are pursued in the next section.
10.4. Uniqueness properties of primary decomposition.
Recall that for ideals I and J of a ring R, (I : J ) = {x ∈ R  xJ ⊂ I }, which
is also an ideal of R. We abbreviate (I : (x)) to (I : x) and ((x) : J ) to (x : J ).
Exercise: Show that for ideals I and J , I ⊂ (I : J ).
Lemma 255. Let q be a pprimary ideal and x ∈ R.
a) If x ∈ q then (q : x) = R.
42It is amusing to note that most dictionaries do not recognize “irredudant” as an English
word, but mathematicians have been using it in this and other contexts for many years. COMMUTATIVE ALGEBRA 163 b) If x ∈ q then (q : x) is pprimary.
/
c) If x ∈ p then (q : x) = q.
/
Proof. a) If x ∈ q then 1(x) = x ⊂ q, so 1 ∈ (q : x). b) If y ∈ (q : x), then xy ∈ q; by
assumption x ∈ q, so y n ∈ q for some n and thus y ∈ r(q) = p. So q ⊂ (q : x) ⊂ p;
/
taking radicals we get r((q : x)) = p. Moreover, if yz ∈ (q : x) with y ∈ (q : x),
/
then xyz = y (xz ) ∈ q, so (xz )n = xn z n ∈ q for some n, and xn ∈ q =⇒ (z m )n ∈ q
/
for some n ∈ Z+ , thus z mn ∈ q ⊂ (q : x).
c) We have in all cases that q ⊂ (q : x). If x ∈ p = r(q) and y ∈ (q : x), then
/
xy ∈ q; since no power of x is q, we must have y ∈ q.
∩n
Theorem 256. (First Uniqueness Theorem) Let I = i=1 qi be any irredundant
primary decomposition of the ideal I . Let pi = r(qi ). Then the pi ’s are precisely the
prime ideals of the form r((I : x)) as x ranges through elements of R. In particular,
they are independent of the choice of irredundant primary decomposition.
∩
∩
Proof. For x ∈ R we have (I : x) = ( i qi : x) = i (qi : x), so
∩
∩
r((I : x)) =
r((qi : x)) =
pj
x∈qj
/ i by Lemma 255. If r(I : x) is prime, then r(I : x) =
pj for some∩ . Conversely, for each i, by irredundancy of the decomposition there
j
exists xi ∈ j ̸=i qj \ qi and then the Lemma implies r(I : xi ) = pi .
∩n
Proposition 257. Let I = i=1 qi be a primary decomposition of an ideal I , with
pi = r(qi ). Then any prime ideal p containing I contains pi for some i.
∩
Proof. If p ⊃ I = i qi , then
∩
∩
p = r(p) ⊃
r(qi ) =
pi ,
i i so by XX, p ⊃ pi for some i.
Thus a minimal element of the set of associated primes of I is precisely a prime
which is minimal over I (or equivalently, a minimal prime of R/I ). This has an
important consequence:
Corollary 258. Let I be a decomposable ideal in a ring R. Then the quotient
ring R/I has only ﬁnitely many minimal primes. Thus a Laskerian ring – and, in
particular, a Noetherian ring – has only ﬁnitely many minimal primes.
In the next section we will give a rather diﬀerent proof of the fact that a Noetherian
ring has only ﬁnitely many minimal primes: we will use topological methods!
Exercise 10.8: Show that an inﬁnite Boolean ring is not Laskerian.
It is therefore unambiguous to speak of a minimal associated prime of I . A
synonym for a minimal associated prime is an isolated prime, whereas a nonisolated prime is often called an embedded prime.43
43It is, alas, not our intention to go deeply enough into theory to justify this terminology. 164 PETE L. CLARK Proposition 259. Let I ⊂ R be a decomposable ideal, I =
primary decomposition, and pi = r(qi ). Then
n
∪
pi = {x ∈ R : (I : x) ̸= I }. ∩n
i=1 qi an irredundant i=1 In particular, if the zero ideal is decomposable, then the set of zero divisors of R is
the union of the minimal primes.
Proof. By passage to the quotient ring R/I , we may assume that I = 0. Let
∩r
0 = i=1 qi be a primary decomposition, with pi = r(qi ). For x ∈ R, ((0) : x) ̸= (0)
iﬀ x is a zerodivisor, so it suﬃces to show the last statement of the proposition,
that the union of the minimal primes is the set of all zerodivisors. Let D be the
set of all zero divisors, so from Exercise 3.X and the proof of Theorem 256 we have
∪
∪∩
∪
D = r (D ) =
r((0 : x)) =
pj ⊂
pj .
0̸=x 0̸=x x∈qj
/ j Conversely, by Theorem 256 each pi is of the form r((0 : x)) for some x ∈ R.
Theorem 260. (Second Uniqueness Theorem) Let I be an ideal of R, and let
n
m
∩
∩
qi = I =
rj
i=1 j =1 be two irredundant primary decompositions for an ideal I . By Theorem 256 we
know that m = n and that there is a reordering r1 , . . . , rn of the rj ’s such that for
1 ≤ i ≤ n, r(qi ) = pi = r(ri ). Moreover, if pi is minimal, then qi = rj .
In other words, the primary ideals corresponding to the minimal primes are independent of the primary decomposition.
We will use the technique of localization to prove this result, so ﬁrst we need
some preliminaries on the eﬀect of localization on a primary decomposition.
Proposition 261. Let R be a ring, S ⊂ R a multiplicatively closed set, and q be a
pprimary ideal. Write ι : R → S −1 R for the localization map.
a) If S ∩ p ̸= ∅, then ι∗ (q) = S −1 R.
b) If S ∩ p = ∅, then ι∗ (q) is ι∗ (p)primary, and ι∗ (ι∗ (q)) = q.
Proof. a) If x ∈ S ∩ p, then for some n ∈ Z+ , xn ∈ S ∩ q, so ι∗ (q) contains a unit
of S −1 R and is therefore S −1 R. Part b) follows immediately from Proposition 165
and Proposition 167a).
∩n
Proposition 262. Let S ⊂ R be a multiplicatively closed set, and let I = i=1 qi
be an irredundant primary decomposition of an ideal I . Put pi = r(qi ) and suppose
that the numbering is such that S ∩ pi = ∅ for i ≤ m and S ∩ pi ̸= ∅ for i > m. Then:
ι∗ (I ) = m
∩ ι∗ (qi ), i=1 ι∗ ι∗ (I ) = m
∩ qi , i=1 and both of these are irrendundant primary decompositions. COMMUTATIVE ALGEBRA 165 Exercise 10.9: Prove Proposition 262.
Proof of Theorem 260: let pi be a minimal associated prime, and put S = R \ pi .
Certainly S is a multiplcatively closed set, and moreover by minimality pi is the
unique associated prime which is disjoint from S . Applying Proposition 262 to both
primary decompositions gives
qi = ι∗ ι∗ (I ) = ri . 10.5. Applications in dimension zero.
We now give the proof of the uniqueness portion of Theorem 217. Let m1 , . . . , mn
be the distinct maximal ideals of the Artinian ring R. As in the proof of Theorem
∏n
217a) there exists k ∈ Z+ such that i=1 mk = ∩n mk = 0. For each i, the radical
i
i=1 i
r(mk ) is the maximal ideal mi , so by Proposition 247 each mk is an mi primary
i
i
∩n
ideal. Thus 0 = i=1 mk is a primary decomposition of the zero ideal which is
i
moreover immediately seen to be irredundant. Since all the primes mi are maximal, the desired uniqueness statement of Theorem 217b) follows from the Second
Uniqueness Theorem (Theorem 260) for primary decompositions.
10.6. Applications in dimension one.
Let R be a onedimensional Noetherian domain, and∩I a nonzero ideal. Then
n
by Theorem 252, I has a primary decomposition: I = i=1 qi , where pi = r(qi ) ⊃
qi ⊃ I is a nonzero prime ideal. But therefore each pi is maximal, so that the pi ’s
are pairwise comaximal. By Proposition 124, so too are the qi ’s, so the Chinese
Remainder Theorem applies to give
I= n
∩ qi = i=1 n
∏ qi , i=1 and
R/I ∼
= n
∏ R/qi . i=1 Thus in this case we can decompose any proper ideal as a ﬁnite product of primary
ideals and not just a ﬁnite intersection. Moreover, for I ̸= 0, all the associated
primes are minimal over I , so the Uniqueness Theorems (Theorems 256 and 260)
simply assert that the ideals qi are unique. This observation will be very useful in
our later study of ideal theory in one dimensional Noetherian domains.
11. Affine k algebras and the Nullstellensatz
Let k be a ﬁeld. By an aﬃne algebra over k we simply mean a ﬁnitely generated
k algebra. Of all the various and sundry classes of commutative rings we have met
and will meet later in these notes, aﬃne algebras are probably the most important
and most heavily studied, because of their connection to algebraic geometry. 166 PETE L. CLARK 11.1. Zariski’s Lemma.
In 1947 the great algebraic geometer Oscar Zariski published a short note [Zar47]
proving the following result.
Theorem 263. (Zariski’s Lemma) Let k be a ﬁeld, A a ﬁnitely generated k algebra
and m a maximal ideal of A. Then A/m is a ﬁnite degree ﬁeld extension of k .
Exercise X.X: Show that the following is an equivalent restatement of Zariski’s
Lemma: let K/k be a ﬁeld extension such that K is ﬁnitely generated as a k algebra. Then K/k is an algebraic ﬁeld extension.
Notwithstanding its innocuous initial appearance, Zariski’s Lemma is a useful result for aﬃne algebras over an aribtrary ﬁeld which, in the special case that k is
algebraically closed, carries all of the content of Hilbert’s Nullstellensatz, the main
theorem of this section.
So how do we prove Zariski’s Lemma?
Oh, let us count the ways! The literature contains many interesting proofs, employing an impressively wide range of ideas and prior technology. We will in fact give
several diﬀerent proofs during the course of these notes. Of course some pride of
place goes to the ﬁrst proof that we give, so after much thought (and after changing
our mind at least once!) we have decided on the following.
Proof of Zariski’s Lemma via the ArtinTate Lemma:
As in Exercise X.X, it suﬃces to prove the following: let K/k be a ﬁeld extension
which is ﬁnitely generated as a k algebra. Then K/k is algebraic.
Well, suppose otherwise: let x1 , . . . , xn be a transcendence basis for K/k (where
n ≥ 1 since K/k is transcendental), put k (x) = k (x1 , . . . , xn ) and consider the
tower of rings
(16) k ⊂ k (x) ⊂ K. To be sure, we recall the deﬁnition of a transcendence basis: the elements xi are
algebraically equivalent over k and K/k (x) is algebraic. But since K is a ﬁnitely generated k algebra, it is certainly a ﬁnitely generated k (x)algebra and thus K/k (x)
is a ﬁnite degree ﬁeld extension. Thus the ArtinTate Lemma applies to (16): we
conclude that k (x)/k is a ﬁnitely generated k algebra. But this is absurd. It implies
the much weaker statement that k (x) = k (x1 , . . . , xn−1 )(xn ) is ﬁnitely generated
as a k (x1 , . . . , xn−1 )[xn ]algebra, or weaker yet, that there exists some ﬁeld F such
that F (t) is ﬁnitely generated as an F [t]algebra: i.e., there exist ﬁnitely many rai (t)
tional functions {ri (t) = pi (t) }N such that every rational function is a polynomial
i=1
q
in the ri ’s with k coeﬃcients. But F [t] is a PID with inﬁnitely many nonassociate
nonzero prime elements q (e.g. adapt Euclid’s argument of the inﬁnitude of the
primes), so we may choose a nonzero prime element q which does not divide qi (t)
for any i. It is then clear that 1 cannot be a polynomial in the ri (t)’s: for instance,
q
evaluation at a root of q in F leads to a contradiction.
Remark: The phenomenon encountered in the endgame of the preceding proof COMMUTATIVE ALGEBRA 167 will be studied in great detail in §12. What we are actually showing is that for any
ﬁeld F , the polynomial ring F [t] is not a Goldman domain, and indeed this is
closely related to the fact that Spec F [t] is inﬁnite. More on this later!
11.2. Hilbert’s Nullstellensatz.
Let k be a ﬁeld, let Rn = k [t1 , . . . , tn ], and write An for k n . We wish to introduce an antitone Galois connection (V, I ) between subsets of Rn and subsets of An .
Namely:
For S ⊂ An , we put
I (S ) = {f ∈ Rn  ∀x ∈ S, f (x) = 0}.
In other words, I (S ) is the set of polynomials which vanish at every element of S .
Conversely, for J ⊂ Rn , we put
V (J ) = {x ∈ An  ∀f ∈ J, f (x) = 0}.
This is nothing else than the Galois relation associated to the relation f (x) = 0 on
the Cartesian product Rn × An .
As usual, we would like to say something about the induced closure operators
on Rn and An . First, for any subset S of An , I (S ) is not just a subset but an ideal
of Rn . In fact I (S ) is a radical ideal: indeed, if f n ∈ I (S ) then f n vanishes on
every point of S , so f vanishes at every point of S .
This little bit of structure pulled from thin air will quicken the heart of any Bourbakiste. But beyond the formalism, the key question is: exactly which sets are
closed? Without knowing this, we haven’t proved the Nullstellensatz any more
than the analogous formalities between sets and groups of automorphisms prove
the Galois correspondence for Galois ﬁeld extensions.
Indeed, an ideal I is radical if f n ∈ I implies f ∈ I . But if f n vanishes identically on S , then so does f .
The closed subsets of An are closed under arbitrary intersections (including the
“empty intersection”: An = V ((0)) and under ﬁnite unions (including the “empty
union”: ∅ = V ({1}) = V (Rn ), and therefore form the closed sets for a unique
topology on An , the Zariski topology.
Exercise: a) Prove these facts.
b) Show that the Zariski topology on An coincides with the topology it inherits as
/k
a subset of Ank .
/
c) Show that the Zariski topology is T1 : i.e., singleton subsets are closed.
d) Show that when n = 1, the Zariski topology is the coarsest T1 topology on k :
namely, the topology in which a proper subset is closed iﬀ it is ﬁnite.
e) For any n ≥ 1, show that the Zariski topology on k n is discrete iﬀ k is ﬁnite.
f) For any inﬁnite ﬁeld and m, n ≥ 1, show that the Zariski topology on k m+n is
strictly ﬁner than the product of the Zariski topologies on k m and k n . 168 PETE L. CLARK Remark: It is often lamented that the Zariski topology (especially when k = C) is
distressingly “coarse”. It is true that it is much coarser than the “analytic topology”
on k n when k is a topological ﬁeld (i.e., the product topology from the topology
on k ). But in fact from an algebraic perspective the Zariski topology is if anything
too ﬁne: we will see why later on when we extend the topology to all prime ideals
on an aﬃne algebra. But also the fact that the Zariski topology on a Fn is discrete
q
creates many geometric problems.
Exercise: Let k be a ﬁeld, n ∈ Z+ as above. Explicitly compute the ideal I (k n ),
i.e., the set of all polynomials which vanish at every point of k n . Do we necessarily
have I (k n ) = {0}? (Hint: No, but we can ﬁgure out exactly for which ﬁelds this
occurs.)
Lemma 264. For a = (a1 , . . . , an ) ∈ k n , put ma = ⟨x1 − a1 , . . . , xn − an ⟩. Then:
a) We have Rn /ma = k . In particular ma is maximal.
b) ma = I ({a}) is the ideal of all functions vanishing at a.
c) The assignment a → ma is a bijection from k n to the set of all maximal ideals
m of Rn such that Rn /m = k .
Proof. Part a) is obvious (but important).
b) Certainly each xi − ai vanishes at a, so ma ⊂ I ({a}). But by part a) ma is a
maximal ideal, whereas 1 ∈ I ({a}), so we must have ma = I ({a}).
/
c) The mapping a → ma is certainly an injection from k n to the set of maximal
ideals with residue ﬁeld k . Conversely, let m be an ideal of Rn such that R/m = k .
For 1 ≤ i ≤ n let ai be the image of xi in R/m = k . Then m ⊃ ma so again we
must have equality.
We now pause for a very important deﬁnition. A ring R is a Jacobson ring if it
is “suﬃciently many maximal ideals”: more precisely, such that every prime ideal
p of R is the intersection of the maximal ideals that contain it.
Exercise X.X: a) Show that a ring R is a Jacobson ring iﬀ for every ideal I , the
intersection of all maximal ideals containing I is rad(I ).
b) Show that every homomorphic image of a Jacobson ring is Jacobson.
Proposition 265. (Rabinowitsch Trick [Ra30]) Let k be any ﬁeld and n ∈ Z+ .
a) The ring Rn = k [x1 , . . . , xn ] is a Jacobson ring.
b) It follows that any aﬃne algebra is a Jacobson ring.
Proof. a) It is suﬃcient to show that for each prime ideal p of Rn and a ∈ R \ p,
there exists a maximal ideal m containing p and not containing a.
1
To show this, put Ra := R[ a ], and let pa = pRa be the pushed forward ideal.
Since p does not meet the multiplicative set generated by a, pa is still prime in
Ra . Let ma be any maximal ideal of Ra conaining pa , and let m = ma ∩ R be its
contraction to R: a priori, this is a prime ideal. There is an induced k algebra
embedding R/m → Ra /ma . But Ra is still a ﬁnitely generated algebra so by
Zariski’s Lemma (Theorem 263) Ra /ma is ﬁnite dimensional as a k vector space,
hence so is the subspace R/m. Thus the domain R/m must be a ﬁeld, a special case
of our coming work on integral extensions. Concretely, let x ∈ (R/m)• and write
out a linear dependence relation of minimal degree among the powers of x:
xn + cn−1 xn−1 + . . . + c1 x + c0 , ci ∈ k, c0 ̸= 0. COMMUTATIVE ALGEBRA 169 Thus (
) −1
x xn−1 + cn−1 xn−2 + . . . + c1 =
,
c0
so x is invertible. Thus m is the desired maximal ideal.
b) This follows immediately from Exercise X.X.
Remark: It seems that “J.L. Rabinowitch”, the author of [Ra30], is the same person as George Yuri Rainich, a distinguished RussoAmerican mathematician of the
early twentieth century.
Here is one last fact we can prove before imposing the hypothesis that k is algebraically closed.
Proposition 266. Let k be any ﬁeld and J an ideal of k [x] = k [x1 , . . . , xn ]. Then:
a) V (J ) = V (rad J ).
b) For any subset S ⊂ k n , I (S ) is a radical ideal.
v) I (V (J )) is a radical ideal containing rad J .
Proof. The underlying mechanism here is the following truly basic observation: for
f ∈ k [x], P ∈ k n and m ∈ Z+ , we have
f (P ) = 0 ⇐⇒ f m (P ) = 0.
a) Since J ⊂ rad J we have V (J ) ⊃ V (rad J ). Conversely, let P ∈ V (J ) and
f ∈ rad J . Then there exists m ∈ Z+ such that f m ∈ J , so f m (P ) = 0 and thus
f (P ) = 0. It follows that P ∈ V (rad J ).
b) Similarly, for any f ∈ k [x] and m ∈ Z+ , if f m ∈ I (S ), then for all P ∈ S ,
f m (P ) = 0. But this implies f (P ) = 0 for all P ∈ S and thus f ∈ I (S ).
c) This follows immediately from parts a) and b) and the tautological fact that for
any ideal J of k [x], I (V (J )) ⊃ J .
Finally we specialize to the case in which the ﬁeld k is algebraically closed. We have
done almost all the work necessary to establish the following fundamental result.
Theorem 267. (Hilbert’s Nullstellensatz) Let k be an algebraically closed ﬁeld, let
k [x] = k [x1 , . . . , xn ]. Then:
a) I induces a bijective correspondence between the singleton sets of k n and the
maximal ideals: a ∈ k n → ma = ⟨x1 − a1 , . . . , xn − an ⟩.
b) For any Zariskiclosed subset S ⊂ k n , V (I (S )) = S .
c) For any ideal J of Rn , I (V (J )) = rad(J ).
Thus there is an inclusionreversing, bijective correspondence between Zariskiclosed
subsets of k n and radical ideals of k [x].
Proof. a) Let m be a maximal ideal of k [x]. By Theorem 263, the residue ﬁeld
k [x]/m is a ﬁnite degree extension of k . Since k is algebraically closed, this forces
k [x]/m = k , and now Lemma 264 applies.
b) There is no content here: it is part of the formalism of Galois connections.
c) By Proposition 266, it is no loss of generality to assume that J is a radical ideal.
Further, by Proposition 265, k [x] is a Jacobson ring, so any radical ideal J is the
intersection of the maximal ideals m containing it. This is true over any ﬁeld k .
But combining with part a), we get that J is an intersection of maximal ideals of
the form ma for certain points a ∈ k n . Since ma = I ({a}), J ⊂ ma iﬀ every element
of J vanishes at a, in other words iﬀ a ∈ V (J ). Thus J is equal to the set of all
polynomials f ∈ Rn which vanish at every point of V (J ): J = I (V (J ))! 170 PETE L. CLARK Exercise: Let k be any ﬁeld. Show that if either part a) or part c) of Theorem 267
holds for the rings k [x1 , . . . , xn ], then k is algebraically closed. (Hint: in fact both
parts fail for each n ∈ Z+ , including n = 1.)
Exercise: Show that Zariski’s Lemma in the case that k is algebraically closed
is equivalent to the following statement: let J = ⟨f1 , . . . , fm ⟩ be an ideal in
k [x1 , . . . , xn ]. Then either there exists a simultaneous zero a of f1 , . . . , fm or there
exist polynomials g1 , . . . , gm such that g1 f1 + . . . + gm fm = 1.
11.3. Other Nullstellens¨tze.
a
11.3.1. The Real Nullstellensatz.
Recall that a ﬁeld k is formally real if it is not possible to express −1 as a
sum of (any ﬁnite number of) squares in k .
Exercise: Let k be a formally real ﬁeld.
a) Show that k is not algebraically closed.
b) Show that any subﬁeld of k is formally real.
A ﬁeld k is real closed if it is formally real and admits no proper formally real
algebraic extension. So for example R is real closed and Q is formally real but not
real closed.
As we saw, even the weak Nullstellensatz fails for polynomial rings over any nonalgebraically closed ﬁeld k . However, when k is formally real one can ﬁnd counterexamples to the Nullstellensatz of a particular form, and when k is real closed
one can show that these counterexamples are in a certain precise sense the only
ones, leading to an identiﬁcation of the closure operation J → I (V (J )) in this case.
In any commutative ring R, an ideal I is real if for all n ∈ Z+ and x1 , . . . , xn ∈ R,
x2 + . . . + x2 ∈ I implies x1 , . . . , xn ∈ I .
n
1
Exercise: Show that a prime ideal p of R is real iﬀ the fraction ﬁeld of R/p is
a formally real ﬁeld.
Exercise: Show that any real ideal is a radical ideal.
So what? The following result gives the connection to the Nullstellensatz.
Proposition 268. Let k be a formally real ﬁeld and k [x] = k [x1 , . . . , xn ]. For any
ideal J of k [x], its closure J = I (V (J )) is a real ideal.
2
2
Proof. Let f1 , . . . , fm ∈ k [x] be such that f1 + . . . + fm ∈ J . Then for any P ∈
2
2
V (J ), we have f1 (P ) + . . . + fm (P ) = 0. Since k is formally real, this implies
f1 (P ) = . . . = fm (P ), and thus f1 , . . . , fm ∈ I (V (J )) = J . 12. Goldman domains and HilbertJacobson rings COMMUTATIVE ALGEBRA 171 12.1. Goldman domains.
Lemma 269. Let R be a domain with fraction ﬁeld K . TFAE:
(i) K is ﬁnitely generated as an Ralgebra.
(ii) There exists f ∈ K such that K = R[f ].
Proof. Of course (ii) =⇒ (i). Conversely, if K = R[f1 , . . . , fn ], then write fi =
1
and then K = R[ q1 ···qn ]. pi
qi , A ring satisfying the conditions of Lemma 269 will be called a Goldman domain.
Exercise: Show that an overring44 of a Goldman domain is a Goldman domain.
Lemma 270. Let R be a domain with fraction ﬁeld K , and 0 ̸= x ∈ R. TFAE:
(i) Any nonzero prime ideal of R contains x.
(ii) Any nonzero ideal contains a power of x.
(iii) K = R[x−1 ].
Proof. (i) =⇒ (ii): let I be a nonzero ideal. If I is disjoint from {xn }, then by
Multiplicative Avoidance (117), I can be extended to a prime ideal disjoint from
{xn }, contradicting (i).
(ii) =⇒ (iii): Let 0 ̸= y ∈ R. By (ii), we have (y ) contains some power of x,
say xk = yz . But this implies that y is a unit in R[x−1 ].
(iii) =⇒ (i): The prime ideals that are killed in the localization map R → R[x−1 ]
are precisely those which meet the multiplicatively closed set {xk }, i.e., contain
x.
Corollary 271. For an integral domain R, TFAE:
(i) R is a Goldman domain.
(ii) The intersection of all nonzero prime ideals of R is nonzero.
Exercise X.X: Prove Corollary 271. Easy examples of Goldman domains: a ﬁeld,
k [[t]], Z(p) . In fact we have developed enough technology to give a remarkably clean
characterization of Noetherian Goldman domains.
Theorem 272. Let R be an integral domain.
a) If R has only ﬁnitely many primes, then R is a Goldman domain.
b) If R is a Noetherian Goldman domain, then R has ﬁnitely many primes.
c) A Noetherian Goldman domain is either a ﬁeld or a onedimensional domain.
Proof. It is harmless to assume rhroughout that R is not a ﬁeld, and we do so.
a) Suppose that R has only ﬁnitely many primes, and let p1 , . . . , pn be the
nonzero prime ideals of R. For 1 ≤ i ≤ n, let 0 ̸= xi ∈ pi , and put x = x1 · · · xn .
Then the multiplicative set S generated by x meets every nonzero prime of R, so
1
that S −1 R has only the zero ideal. In other words, R[ x ] is the fraction ﬁeld of R,
so R is a Goldman domain. (Alternately, this is an almost immediate consequence
of Corollary 271.)
1
b) Similarly, for a Goldman domain R we can write K = R[ x ] for x ∈ R and
then every nonzero prime of R contains x. Suppose ﬁrst that (x) itself is prime,
necessarily of height one by the Hauptidealsatz, hence if R has any primes other
than (0) and (x) – especially, if it has inﬁnitely many primes – then it has a height
44An overring of a domain R is a ring intermediate between R and its fraction ﬁeld K . 172 PETE L. CLARK two prime q. But by Corollary 226 a Noetherian ring cannot have a height two
prime unless it has inﬁnitely many height one primes, a contradiction. So we may
assume that (x) is not prime, and then the minimal primes of the Noetherian ring
R/(x) are ﬁnite in number – say p1 , . . . , pn – and correspond to the primes of R
which are minimal over x, so again by the Hauptidealsatz they all have height one.
Similarly, if R has inﬁnitely many primes there would be, for at least one i (say
i = 1), a height two prime q ⊃ p1 . But then by Corollary 226 the “interval” (0, q)
is inﬁnite. Each element of this set is a height one prime ideal containing (x), i.e.,
is one of the pi ’s, a contradiction. Finally, part c) follows by once again applying
Corollary 226: a Noetherian ring of dimension at least two must have inﬁnitely
many primes.
Remark: a nonNoetherian Goldman domain can indeed have inﬁnitely many primes
and/or primes of arbitrarily large height.
Proposition 273. Let R be an integral domain. Then the polynomial ring R[t] is
not a Goldman domain.
Proof. Let K be the fraction ﬁeld of R. If R[t] is a Goldman domain, then by
Exercise XX, so is its overring K [t]. But the ring K [t] is a Noetherian domain with
inﬁnitely many primes – for instance, Euclid’s proof of the inﬁnitude of primes in
Z carries over verbatim to K [t] – so Theorem 272 applies to show that K [t] is not
a Goldman domain.
Proposition 274. Let R be a domain, and T ⊃ R an extension domain which is
algebraic and ﬁnitely generated as an Ralgebra. Then R is a Goldman domain iﬀ
T is a Goldman domain.
Proof. Let K and L be the fraction ﬁelds of R and T , respectively. Suppose ﬁrst
1
1
that R is a Goldman domain: say K = R[ u ]. Then T [ u ] is algebraic over the
1
ﬁeld K , so is a ﬁeld, hence we have L = T [ u ]. Conversely, suppose that T is
1
a Goldman domain: say L = T [ v ]; also write T = R[x1 , . . . , xk ]. The elements
v −1 , x1 , . . . , xk are algebraic over R hence satisfy polynomial equations with coeﬃcients in R. Let a be the leading coeﬃcient of a polynomial equation for v −1 and
b1 , . . . , bk be the leading coeﬃcients of polynomial equations for x1 , . . . , xk . Let
R1 := R[a−1 , b−1 , . . . , b−1 ]. Now L is generated over R1 by x1 , . . . , xk , v −1 , all of
1
k
which are integral over R1 . Therefore L is integral over R1 . Since L is a ﬁeld,
it follows that R1 is a ﬁeld, necessarily equal to K , and this shows that R is a
Goldman domain.
Corollary 275. Let R ⊂ S be an inclusion of domains, with R a Goldman domain.
Suppose that u ∈ S is such that R[u] is a Goldman domain. Then u is algebraic
over R, and R is a Goldman domain.
Theorem 276. For an integral domain R, TFAE:
(i) R is a Goldman domain.
(ii) There exists a maximal ideal m of R[t] such that m ∩ R = (0).
1
Proof. (i) =⇒ (ii): We may assume WLOG that R is not a ﬁeld. Write K = R[ u ].
1
Deﬁne a homomorphism φ : R[t] → K by sending t → u . Evidently φ is surjective,
so its kernel m is a maximal ideal, and clearly we have m ∩ R = 0.
(ii) =⇒ (i): Suppose m is a maximal ideal of R[t] such that m ∩ R = (0). Let COMMUTATIVE ALGEBRA 173 v be the image of t under the natural homomorphism R[t] → R[t]/m. Then R[v ] is
a ﬁeld, so by Corollary 275, R is a Goldman domain.
We deﬁne a prime ideal p of a ring R to be a Goldman ideal if R/p is a Goldman
domain. Write G Spec R for the set of all Goldman ideals. Thus a Goldman ideal
is more general than a maximal ideal but much more special than a prime ideal.
Proposition 277. Let R be a ring and I an ideal of R.
a) The nilradical of R is the intersection of all Goldman ideals of R.
b) The radical of I is the intersection of all Goldman ideals containing I .
∩
∩
Proof. a) We know that N = p∈Spec R p, so certainly N ⊂ p∈G Spec R p. Conversely, suppose x ∈ R \ N . The ideal (0) is then disjoint from the multiplicative set
S = {xn }. By multiplicative avoidance, we can extend (0) to an ideal p maximal
with respect to disjointness from S . We showed earlier that p is prime; we now
claim that it is a Goldman ideal. Indeed, let x denote the image of x in R = R/p.
By maximality of p, every nonzero prime of R contains x. By Lemma 270, this implies R[x−1 ] is a ﬁeld, thus R is a Goldman domain, and therefore p is a Goldman
ideal which does not contain x. Part b) follows by correspondence, as usual.
The following result may seem completely abstruse at the moment, but soon enough
it will turn out to be the key:
Corollary 278. An ideal I in a ring R is a Goldman ideal iﬀ it is the contraction
of a maximal ideal in the polynomial ring R[t].
Proof. This follows from Theorem 276 by applying the correspondence principle to
the quotient ring R/I .
Theorem 279. a) Let M be a maximal ideal in R[t], and suppose that its contraction m = M ∩ R is maximal in R. Then M can be generated by m and by
one additional element f , which can be taken to be a monic polynomial which maps
modulo m to an irreducible polynomial in R/m[t].
b) If, moreover, we suppose that R/m is algebraically closed, then M = ⟨m, t − a⟩
for some a ∈ R.
Proof. a) Since M contains m, by correspondence M may be viewed as a maximal
ideal of R[t]/mR[t] ∼ (R/m)[t], a PID, so corresponds to an irreducible polynomial
=
f ∈ R/m[t]. If f is any lift of f to R[t], then M = ⟨m, f ⟩. Part b) follows
immediately from the observation that an irreducible univariate polynomial over
an algebraically closed ﬁeld is linear.
The following more elementary result covers the other extreme.
Theorem 280. Let R be a domain, with fraction ﬁeld K . Let ι : R[t] → K [t] be
the natural inclusion. Then ι∗ induces a bijection between the prime ideals P of
R[t] such that P ∩ R = {0} and the prime ideals of K [t].
Proof. Let S = R \{0}. The key observation is that S −1 R[t] = K [t]. Recall (Proposition 167) that in any localization map R → S −1 R, the prime ideals which push
forward to the unit ideal are precisely those which meet S , whereas the localization
map restricted to all other prime ideals is a bijection onto the set of prime ideals
of S −1 R. Applying that in this case gives the desired result immediately! 174 PETE L. CLARK 12.2. Hilbert rings.
To put Theorem 279 to good use, we need to have a class of rings for which the
contraction of a maximal ideal from a polynomial ring is again a maximal ideal. It
turns out that the following is the right class of rings:
Deﬁnition: A Hilbert ring is a ring in which every Goldman ideal is maximal.
Proposition 281. Any quotient ring of a Hilbert ring is a Hilbert ring.
Proof. This follows immediately from the correspondence between ideals of R/I
and ideals of R containing I .
A direct consequence of the deﬁnition and Proposition 277 is the following:
Proposition 282. Let I be an ideal in a Hilbert ring R. Then the intersection
∩
m sup I m of all maximal ideals m containing I is rad(I ).
Examples: Any zero dimensional ring is a Hilbert ring: in this case all prime ideals
are maximal. Especially, a ﬁeld is a Hilbert ring, as is any Artinian ring, or any
Boolean ring.
Exercise: a) Let R be a onedimensional Noetherian domain. TFAE:
(i) R is a Hilbert ring.
(ii) The Jacobson radical of R is 0.
(iii) R has inﬁnitely many prime ideals.
(iv) R is not a Goldman domain.
b) Deduce that the ring Z of integers is a Hilbert domain.
Theorem 283. Let R be a Hilbert ring, and S a ﬁnitely generated Ralgebra. Then:
a) S is also a Hilbert ring.
b) For every maximal ideal P of S , p := P ∩ R is a maximal ideal of R.
c) The degree [S/P : R/p] is ﬁnite.
Proof. a) It suﬃces to show that R is a Hilbert ring iﬀ R[t] is a Hilbert ring, for
then, if R is a Hilbert ring, by induction any polynomial ring R[t1 , . . . , tn ] is a
Hilbert ring, and any ﬁnitely generated Ralgebra is a quotient of R[t1 , . . . , tn ] and
thus a Hilbert ring. Note also that since R is a homomorphic image of R[t], if R[t]
is a Hilbert domain, then so also is R.
So suppose that R is a Hilbert ring, and let q be a Goldman ideal in R[t]; we must
show that q is maximal. Put p = q ∩ R. As above, we can easily reduce to the case
p = 0, so that in particular R is a domain. Let a be the image of t in the natural
homomorphism R[t] → R[t]/q. Then R[a] is a Goldman domain. By Corollary 275,
a is algebraic over R and R is a Goldman domain. But since we assumed that R
was a Hilbert ring, this means that R is a ﬁeld, and thus R[a] = R[t]/q is a ﬁeld,
so q is maximal.
b) We may write S = R[t1 , . . . , tn ]/I . A maximal ideal m of S is just a maximal
ideal of R[t1 , . . . , tn ] containing I . By Corollary 278, the contraction m′ of m to
R[t1 , . . . , tn−1 ] is a Goldman ideal of the Hilbert ring R[t1 , . . . , tn−1 ], so is therefore
maximal. Moreover, by Theorem 279, m is generated by m′ and an irreducible
polynomial in R/m′ [t], so that the residual extension R[t1 , . . . , tn ]/m has ﬁnite
degree over R[t1 , . . . , tn−1 /m′ . Again, induction gives the full result. COMMUTATIVE ALGEBRA 175 Applying Theorem 283c) in the case R = k is a ﬁeld, we deduce our second proof
of Zariski’s Lemma (Lemma 263).
Moreover, we deduce a strong form of the Dimension Theorem for Noetherian
Hilbert rings.
Theorem 284. Let R be a Noetherian Hilbert ring. Then
dim(R[t]) = dim R + 1.
Proof. Let 0 = p0
p1
...
pd be a chain of prime ideals in R. Then, with
ι : R → R[t] the natural inclusion,
ι∗ p0 ... ι∗ pn ⟨ι∗ (pn ), t⟩ is a chain of prime ideals of R[t] of length d + 1, hence for any ring R we have
dim R[t] ≥ dim R +1. Conversely, it suﬃces to show that the height of any maximal
ideal P of R[t] is at most d + 1. For this, put p = P ∩ R. By Theorem 283,
p is maximal in R, so Theorem 279 tells us that there exists f ∈ R[t] such that
P = ⟨ι∗ p, f ⟩. Applying Krull’s Hauptidealsatz in the quotient ring R[t]/ι∗ p, we get
that the height of P is at most one more than the height of p.
Corollary 285. Let k be a ﬁeld, and put R = k [t1 , . . . , tn ].
a) Then every maximal ideal of R has height n and can be generated by n elements
(and no fewer, by Theorem 228).
b) In particular, dim R = n.
Exercise: Prove Corollary 285.
12.3. Jacobson Rings.
We deﬁne a Jacobson ring to be a ring which satisﬁes the conclusion of Proposition 282, i.e., a ring R in which for every ideal I , the radical of I is the intersection
of all the maximal ideals containing I .
Exercise: Show that for a ring R, TFAE:
(i) R is a Jacobson ring.
(ii) Every prime ideal of R is an intersection of maximal ideals.
(iii) Every radical ideal of R is an intersection of maximal ideals.
Exercise: Show that any quotient of a Jacobson ring is a Jacobson ring.
12.4. HilbertJacobson Rings.
Proposition 286. Suppose R is both a Goldman domain and a Jacobson ring.
Then R is a ﬁeld.
Proof. Let K be the fraction ﬁeld of R, and suppose for a conradiction that R ̸= K .
Then there exists a nonzero nonunit f ∈ R such that K is the localization of R at
the multiplicative subset S = {f, f 2 , . . .}. Let m be a maximal ideal of R. Since R
is not a ﬁeld, m is not zero, and thus the pushforward of R to S −1 R is the unit ideal.
By Prop. X.X, this implies that m meets S . Since m is prime, we conclude f ∈ m.
It follows that the Jacobson radical of R contains f is accordingly nonzero. On the
other hand R, being a domain, has zero nilradical. Thus R is not Jacobson. 176 PETE L. CLARK Our previous work now immediately gives the following result:
Theorem 287. For a commutative ring R, TFAE:
(i) R is a Hilbert ring.
(ii) R is a Jacobson ring.
(iii) For all maximal ideals m of R[t], m ∩ R is a maximal ideal of R.
Proof. (i) =⇒ (ii) by Proposition 282.
(ii) =⇒ (i): Suppose R is Jacobson and p is a Goldman ideal of R. By Exercise
X.X, the Goldman domain R/p is also Jacobson, hence by Proposition 286 R/p is
a ﬁeld, so p is maximal.
(ii) =⇒ (iii) is Theorem 283b).
(iii) =⇒ (i): Suppose R is a ring such that every maximal ideal of R[t] contracts
to a maximal ideal of R, and let p be a Goldman ideal of R. By Corollary 278, p is
the contraction of a maximal ideal of R[t], hence by assumption p is maximal.
In the sequel we will use the consolidated terminology HilbertJacobson ring for
a ring satisfying the equivalent conditions of Theorem 287.
13. The spectrum of a ring
13.1. The Zariski spectrum.
For a commutative ring R, we denote the set of all prime ideals of R by Spec R.
Moreover, we refer to Spec R as the Zariski spectrum – or prime spectrum –
of the ring R.
It is important to notice that Spec R comes with additional structure. First, it
has a natural partial ordering, in which the maximal elements are the maximal
ideals, and the minimal elements are (by deﬁnition) the minimal primes. Also,
as O. Zariski ﬁrst observed, Spec R can be endowed with a topology. To see this,
for any ideal I of R, put V (I ) = {p ∈ Spec R  p ⊃ I }.
Proposition 288. For ideals I and J of R, we have V (I ) = V (J ) iﬀ rad I = rad J .
Proof. For any ideal I and any prime ideal p, p ⊃ I iﬀ p ⊃ rad I , and therefore V (I ) = V (rad I ). Conversely, if V (I ) = V (J ), then the set of prime ideals
containing I is the same as the set of prime ideals containing J . So
∩
∩
rad I =
p=
p = rad J.
p⊃I p⊃J Exercise X.X: Show that I ⊂ J iﬀ V (I ) ⊃ V (J ).
Now we claim that the family of subsets V (I ) of Spec R has the following properties:
(ZT1) ∅ = V (R), Spec R = V ((0)).
∩
(ZT2) If {Ii } is any collection of ideals of R, then i V (Ii ) = V (⟨Ii ⟩). ∩
∪n
n
(ZT3) If I1 , . . . , In are ideals of R, then i=1 V (Ii ) = V (I1 · · · In ) = V ( i=1 Ii ). COMMUTATIVE ALGEBRA 177 ∩
(ZT1) is obvious. As for (ZT2), let p be a prime ideal of R. Then p ∈ i V (Ii ) for
all i iﬀ p ⊃ Ii for all i iﬀ p contains the ideal generated by all Ii . As for (ZT3), p
contains a product of ideals iﬀ it contains one of the ideals of the product.
Therefore there is a unique topology on Spec R in which the closed sets are precisely
those of the form V (I ). This is called the Zariski topology.
It is of course natural to ask for a characterization of the open sets. Recall that a
base for the open sets of a topology is a collection {Bi } of open sets such that:
(BT1) for any point x ∈ Bi ∩ Bj , there exists a k such that x ∈ Bk ⊂ Bi ∩ Bj ;
(BT2) every open set is a union of the Bi ’s contained in it.
For f ∈ R, we deﬁne U (f ) := Spec R \ V ((f )). In other words, U (f ) is the
collection of all prime ideals which do not contain the element f . For f, g ∈ R,
U (f ) ∩ U (g ) is the set of prime ideals p containing neither f nor g ; since p is prime,
this is equivalent to p not containing f g , thus
U (f ) ∩ U (g ) = U (f g ),
which is a stronger property than (BT1). Moreover, any open set U is of ∩ form
the
Spec R \ V (I ). Each ideal I is the union of all of its elements fi , so V (I ) = i V (fi ),
so that
∩
∪
∪
U = Spec R \ V (I ) = Spec R \ V (fi ) = (Spec R \ V (fi )) =
U (fi ).
i i i Proposition 289. Let R be any ring, and consider the canonical homomorphism
f : R → Rred = R/ nil(A). Then f −1 : Spec Rred → Spec R is a homeomorphism.
Exercise X.X: Prove Proposition 289.
Exercise: Let R1 , . . . , Rn be ﬁnitely many rings. Show that Spec(R1 × . . . × Rn ) is
⨿n
canonically homeomorphic to the topological space i=1 Spec Ri .
Exercise: Let R be a Boolean ring. Earlier we deﬁned a topology on the set
“M (R)” of all maximal ideals of R. But, as we know, a Boolean ring all prime
ideals are maximal, so as sets M (R) = Spec R. Show that moreover the topology
we deﬁned on M (R) is the Zariski topology on Spec R.
13.2. Properties of the spectrum: quasicompactness.
More than sixty years ago now, N. Bourbaki introduced the term quasicompact
for a topological space X for which any open covering has a ﬁnite subcovering.
The point of this terminology is to reserve compact for a space which is both
quasicompact and Hausdorﬀ, and thus emphasize that most of the nice properties
of compact spaces in classical topology do rely on the Hausdorﬀ axiom. Nowhere
is this terminology more appropriate than in the class of spectral spaces, which
as we have seen above, are only Hausdorﬀ in the comparatively trivial case of a
zerodimensional ring. On the other hand:
Proposition 290. For any commutative ring R, Spec R is quasicompact. 178 PETE L. CLARK Proof. Let {Ui } be any open covering of Spec R. For each p ∈ Spec R, there
exists an element U of the cover containing p, and thus a principal open set X (f )
containing p and contained in U . Therefore there is a reﬁnement of the cover
consisting of principal open subsets, and if this reﬁnement has a ﬁnite cover, then
the original cover certainly does as well. Thus it suﬃces to assume that the Ui ’s
∪
are basic open sets.45 So now suppose that Spec R = i X (fi ). Then we have
∪
∪
∩
Spec R =
X (fi ) = (Spec R \ V (fi )) = Spec R \
V (fi ),
i i i ∩
so that ∅ = i V (fi ) = V (⟨fi ⟩). Therefore the ideal I = ⟨fi ⟩ contains 1, and this
means∩
that there is some ﬁnite subset f1 , . . . , fn of I such that ⟨f1 , . . . , fn ⟩ = R.
∪n
n
Thus i=1 V (fi ) = ∅, or equivalently, Spec R = i=1 X (fi ).
13.3. Properties of the spectrum: separation and specialization.
For the reader’s convenience we brieﬂy recall the “lower” separation axioms:
A topological space X is Kolmogorov – or T0 – if for any distinct points x, y ∈ X ,
the system of neighborhoods Nx and Ny do not coincide. In plainer language,
either there exists an open set U containing x and not containing y , or conversely.
A topological space X is separated – or T1 – if for any distinct points x, y ∈ X ,
there exists both an open set U containing x and not y and an open set V containing y and not x. A space is separated iﬀ all singleton sets {x} are closed iﬀ for all
∩
x ∈ X , U ∈Nx U = {x}.
A topological space X is Hausdorﬀ – or T2 – if for any distinct points x, y ∈ X ,
there exist open neighborhoods U of x and V of y with U ∩ V = ∅. A space is
Hausdorﬀ iﬀ for all x ∈ X , the intersection of all closed neighborhoods of x is {x}.
Easily Hausdorﬀ implies separated implies Kolmogorov. In a general topology
course one learns that neither of the converse implications holds in general. On
the other hand most of the spaces one encounters in analysis and geometry are
Hausdorﬀ, and certainly are if they are Kolmogorov (see Exercise X.X for a nonKolmogorov space arising naturally in analysis). We are about to see that yet a
third state of aﬀairs transpires when we restrict attention to spectra of rings.
Let X be a topological space. We deﬁne a relation → on X by decreeing that
for x, y ∈ X , x → y iﬀ y lies in the closure of the singleton set {x}. This relation
is called specialization, and we read x → y as “x specializes to y ”.
The reader who is familiar with topology but has not seen the specialization relation before will ﬁnd an explanation in part f) of the following exercise.
Exercise X.X:
a) Show that x → y iﬀ Nx ⊂ Ny .
45This is just the familiar, and easy, fact that it suﬃces to verify quasicompactness on any
base for the topology. It is also true, but deeper, that one can verify quasicompactness on any
subbase: Alexander’s Subbase Theorem. COMMUTATIVE ALGEBRA 179 b) Show that specialization satisﬁes the following properties:
(i) reﬂexivity: x → x; (ii) transitivity x → y, y → z =⇒ x → z .
A relation R with these properties is called a quasiordering. Note that a
partial ordering is a quasiordering with the additional axiom of antisymmetry:
xRy, yRx =⇒ x = y .
c) Show that the specialization relation on X is a partial ordering iﬀ X is Kolmogorov.
d) Show that a point y is closed46 iﬀ y → x =⇒ x = y .
e) A point x for which x → y holds for all y ∈ X is called generic. Give an example
of a topological space in which every point is generic.
f) Show that X is separated iﬀ x → y =⇒ x = y .
Exercise X.X: Let X be any set endowed with a quasiordering R. Deﬁne a new
relation x ≡ y if x R y and y R x.
a) Show that ≡ is an equivalence relation on X .
b) Write X ′ for the set of ≡ equivalence classes, and let q : X → X ′ be the natural
map – i.e., x → {y ∈ X  y ≡ x}. Show that the relation R descends to a relation
≤ on X ′ : i.e., for s1 , s2 ∈ X ′ , then by choosing x1 ∈ s1 , x2 ∈ s2 and putting
s1 ≤ s2 ⇐⇒ x1 R x2 ,
the relation ≤ is welldeﬁned independent of the choices of x1 and x2 . Show that
moreover ≤ is a partial ordering on X ′ .
c) Let X be a topological space and R be the specialization relation. Endowing
X ′ with the quotient topology via q , show that the induced relation ≤ on X ′ is
the specialization relation on X ′ , and accordingly by the previous exercise X ′ is a
Kolmogorov space. If it pleases you, show that q : X → X ′ is universal for maps
from X into a Kolmogorov space Y , hence X ′ (or rather, q : X → X ′ ) can be
regarded as the Kolmogorov quotient of X .
Exercise: Let (X, µ) be a measure space, and let L1 be the space of all measurable
∫
∫
functions f : X → R with X f dµ < ∞. For f ∈ L1 , deﬁne f  := X f dµ, and
for ϵ > 0, put B (f, ϵ) = {g ∈ L1  g − f  < ϵ}. Show that the B (f, ϵ)’s form a
base for a topology on L1 , but that this topology is, in general, not Kolmogorov.
Show that the Kolmogorov quotient is precisely the usual Lebesgue space L1 , whose
elements are not functions but classes of functions modulo µ a.e. equivalence.
Proposition 291. For any ring R, the spectrum Spec R is a Kolmogorov space.
Indeed, for prime ideals p, q of R, we have p → q iﬀ q ⊃ p, i.e., the specialization
relation is precisely the opposite relation to the containment of prime ideals.
Proof. For prime ideals p and q we have
p → q ⇐⇒ q ∈ {p} = {f ∈ Spec R  f ⊃ p} ⇐⇒ q ⊃ p.
Thus the specialization relation is just reverse containment of ideals, which certainly
satisﬁes antisymmetry: q ⊂ p, p ⊂ q =⇒ p = q. Now apply Exercise X.Xc). 46Strictly speaking we mean that {y } is closed, but this abuse of terminology is very common
and convenient. 180 PETE L. CLARK Lemma 292. For a ring R, TFAE:
(i) R/ nil R is absolutely ﬂat, i.e., every R/ nil Rmodule is ﬂat.
(ii) R has dimension zero: every prime ideal of R is maximal.
Proof. We use the characterization of absolutely ﬂat rings given in Proposition 180:
it is necessary and suﬃcient that the localization at every maximal ideal be a ﬁeld.
Since the Krull dimension of Rm is the height of R, evidently any absolutely ﬂat
ring has Krull dimension 0. Since Spec(R/ nil R) = Spec R, dim R/ nil R = dim R,
thus (i) =⇒ (ii). Conversely, localizations of reduced rings are reduced, and a
reduced local ring of dimension zero is a ﬁeld, so (ii) =⇒ (i).
Theorem 293. For a commutative ring R, TFAE:
(i) R/ nil R is absolutely ﬂat, i.e., every R/ nil Rmodule is ﬂat.
(ii) R has Krull dimension zero.
(iii) Spec R is a separated space.
(iv) Spec R is a Hausdorﬀ space.
(v) Spec R is a Boolean space.
Proof. The equivalence of (i) and (ii) is Lemma 292.
(ii) ⇐⇒ (iii): a space is separated iﬀ all of its singleton sets are closed. But if p
is a prime ideal, then V (p) consists of all prime ideals containing p, so V (p) = {p}
iﬀ p is maximal.
Certainly (v) =⇒ (iv) =⇒ (iii).
(i) =⇒ (v): Since Spec R = Spec(R/ nil R), we may well assume that R itself
is absolutely ﬂat. Let p and q be distinct prime ideals; since both are maximal,
there exists an element f ∈ p \ q. By Proposition 98, there exists an idempotent
e with (e) = (f ), and therefore e ∈ p \ q. Then D(1 − e), D(e) is a separation of
Spec R. More precisely, D(e) ∩ D(1 − e) = D(e(1 − e)) = D(e − e2 ) = D(0) = ∅,
whereas for any prime ideal p, since 0 = e(1 − e) ∈ p, we must have e ∈ p or
1 − e ∈ pp. Moreover by construction we have p ∈ D(1 − e), q ∈ D(e). This shows
that Spec R is Hausdorﬀ, but in fact it shows more: given any two distinct points
⨿
P ̸= Q of the space X , we found a separation X = U V with P ∈ U, Q ∈ V ,
so that X is zerodimensional. According to Proposition 290, every ring has quasicompact spectrum, so Spec R is Hausdorﬀ, zerodimensional and quasicompact,
i.e., Boolean.
Exercise X.X:
a) Let R be an arbitrary product of ﬁelds. Show that Spec R is a Boolean space.
b) Let {Ri }i∈I be a family of rings, each of which has Krull dimension 0, and put
∏
R = i Ri . Must Spec R be Boolean?
13.4. Irreducible spaces.
A topological space is irreducible if it is nonempty and if it cannot be expressed
as the union of two proper closed subsets.
Exercise X.X: Show that for a Hausdorﬀ topological space X , TFAE:
(i) X is irreducible.
(ii) #X = 1.
Proposition 294. For a topological space X , TFAE:
(i) X is irreducible. COMMUTATIVE ALGEBRA 181 (ii) Every ﬁnite intersection of nonempty open subsets (including the empty intersection!) is nonempty.
(iii) Every nonempty open subset of X is dense.
(iv) Every open subset of X is connected.
Exercise: Prove Proposition 294.
Proposition 295. Let X be a nonempty topological space.
a) If X is irreducible, every nonempty open subset of X is irreducible.
b) If a subset Y of X is irreducible, so is its closure Y .
c) If {Ui } is an open covering of X such that Ui ∩ Uj ̸= ∅ for all i, j and each Ui
is irreducible, then X is irreducible.
d) If f : X → Y is continuous and X is irreducible, then f (X ) is irreducible in Y .
Proof. a) Let U be a nonempty open subset of X . By Proposition 294, it suﬃces
to show that any nonempty open subset V of U is dense. But V is also a nonempty
open subset of the irreducible space X .
b) Suppose Y = A ∪ B where A and B are each proper closed subsets of Y ; since
Y is itself closed, A and B are closed in X , and then Y = (Y ∩ A) ∪ (Y ∩ B ). If
Y ∩ A = Y then Y ⊂ A and hence Y ⊂ A = A, contradiction. So A is proper in Y
and similarly so is B , thus Y is not irreducible.
c) Let V be a nonempty open subset of X . Since the Ui ’s are a covering of X , there
exists at least one i such that V ∩ Ui ̸= ∅, and thus by irreducibility V ∩ Ui is a
dense open subset of Ui . Therefore, for any index j , V ∩ Ui intersects the nonempty
open subset Uj ∩ Ui , so in particular V intersects every element Uj of the covering.
Thus for all sets Ui in an open covering, V ∩ Ui is dense in Ui , so V is dense in X .
d) If f (X ) is not irreducible, there exist closed subsets A and B of Y such that
A ∩ f (X ) and B ∩ f (X ) are both proper subsets of f (X ) and f (X ) ⊂ A ∪ B . Then
f −1 (A) and f −1 (B ) are proper closed subsets of X whose union is all of X .
Let x be a point of a topological space, and consider the set of all irreducible subspaces of X containing x. (Since {x} itself is irreducible, this set is nonempty.) The
union of a chain of irreducible subspaces being irreducible, Zorn’s Lemma says that
there exists at least one maximal irreducible subset containing x. A maximal irreducible subset (which, by the above, is necessarily closed) is called an irreducible
component of X . Since irreducible subsets are connected, each irreducible component lies in a unique connected component, and each connected component is
the union of its irreducible components.
However, unlike connected components, it is possible for a given point to lie in
more than one irreducible component. We will see examples shortly.
In the case of a Zariski topology Spec R, there is an important algebraic interpretation of the irreducible components. Namely, the irreducible components Y of
Spec R correspond to V (p) where p ranges through the minimal primes.
Indeed, we ﬁrst claim that a closed subset V (I ) is irreducible iﬀ rad I = p is prime.
First, if rad I = p is prime, then by Proposition 288 V (I ) = V (p), so that it suﬃces
to show that V (p) is irreducible. If not, there exist ideals I and J such that V (I )
and V (J ) are both proper subsets of V (p) and V (p) = V (I ) ∪ V (J ) = V (IJ ). But
then p = rad(IJ ) ⊃ IJ and since p is prime this implies either p ⊃ I or p ⊃ J . 182 PETE L. CLARK WLOG, suppose p ⊃ I ; then V (p) ⊂ V (I ), so that V (I ) is not proper in V (p),
contradiction.
Next, suppose that V (I ) is irreducible, and suppose that ab ∈ rad(I ). If neither
a nor b is in rad(I ), then V ((a)), V ((b)) do not contain V (I ), but V (a) ∪ V (b) =
V (ab) ⊃ V (I ). Therefore V (a) ∩ V (I ) = V (aI ) and V (b) ∩ V (I ) = V (bI ) are two
proper closed subsets of V (I ) whose union is V (I ), thus V (I ) is reducible.
It follows that the irreducible components – i.e., the maximal irreducible subsets –
are precisely the sets of the form V (p) as p ranges over the distinct minimal prime
ideals. Note that we can now deduce that minimal prime ideals exist in any ring as
a special case of the existence of irreducible components in any topological space,
an example of the use of topological methods to prove purely algebraic results.47
Proposition 296. For any commutative ring, the map p → V (p) gives a bijection
from Spec R to the set of irreducible closed subsets of Spec R.
Exercise X.X: Prove Proposition 296.
Exercise: Explain why Proposition 296 is, in some sense, a Nullstellensatz for an
arbitrary commutative ring.
13.5. Noetherian spaces.
We wish to introduce a property of topological spaces which, from the standpoint
of conventional geometry, looks completely bizarre:
Proposition 297. For a topological space X , TFAE:
(i) Every ascending chain of open subsets is eventually constant.
(ibis) Every descending chain of closed subsets is eventually constant.
(ii) Every nonempty family of open subsets has a maximal element.
(iibis) Every nonempty family of closed subsets has a minimal element.
(iii) Every open subset is quasicompact.
(iv) Every subset is quasicompact.
A space satisfying any (and hence all) of these conditions is called Noetherian.
Proof. The equivalence of (i) and (ibis), and of (ii) and (iibis) is immediate from
taking complements. The equivalence of (i) and (ii) is a general property of partially
ordered sets discussed in section X.X above.
(i) ⇐⇒ (iii): Assume (i), let U be any open set in X and let {Vj } be an open
covering of U . We assume for a contradiction that there is no ﬁnite subcovering.
Choose any j1 and put U1 := Vj1 . Since U1 ̸= U , there exists j2 such that U1 does
not contain Vj2 , and put U2 = U1 ∪ Vj2 . Again our assumpion implies that U2 U ,
and continuing in this fashion we will construct an inﬁnite properly ascending chain
of open subsets of X , contradiction. Conversely, assume (iii) and let {Ui }∞ be an
i=1
∪
inﬁnite properly ascending chain of subsets. Then U = i Ui is not quasicompact.
Obviously (iv) =⇒ (iii), so ﬁnally we will show that (iii) =⇒ (iv). Suppose
that Y ⊂ X is not quasicompact, and let {Vi }i∈I be a covering of Y by relatively
open subsets without a ﬁnite subcover. We may write each Vi as Ui ∩ Y with Ui open
47In this case, the same Zorn’s Lemma argument establishes the existence of minimal primes,
so the topology is not making anything essentially easier for us...yet. COMMUTATIVE ALGEBRA 183 ∪
in Y . Put U = i Ui . Then, since U is quasicompact, there exists a ∪
ﬁnite subset
∪
∪
J ⊂ I such that U = j ∈J Uj , and then Y = U ∩ Y = j ∈J Uj ∩ Y = j ∈J Vj .
Corollary 298. A Noetherian Hausdorﬀ space is ﬁnite.
Proof. In a Hausdorﬀ space every quasicompact subset is closed. Therefore, using
the equivalence (i) ⇐⇒ (iv) in Proposition 297, in a Noetherian Hausdorﬀ space
every subset is closed, so such a space is discrete. But it is also quasicompact, so
it is ﬁnite.
Proposition 299. For a ring R, TFAE:
(i) R satisﬁes the ascending chain condition on radical ideals.
(ii) Spec R is a Noetherian space.
In particular if R, or even Rred = R/ nil(R), is a Noetherian ring, Spec R is a
Noetherian space.
Proof. Since I → V (I ) gives a bijection between radical ideals and Zariski closed
subsets, (ACC) on radical ideals is equivalent to (DCC) on closed subsets. Evidently
these conditions occur if R is itself Noetherian, or, since Spec R is canonically
homeomorphic to Spec Rred , if Rred is Noetherian.
Proposition 300. Let X be a Noetherian topological space.
∪n
a) There are ﬁnitely many closed irreducible subsets {Ai }n such that X = i=1 Ai .
i=1
b) Starting with any ﬁnite family {Ai }n as in part a) and eliminating all reduni=1
dant sets – i.e., all Ai such that Ai ⊂ Aj for some j ̸= i – we arrive at the
set of irreducible components of X . In particular, the irreducible components of a
Noetherian space are ﬁnite in number.
Proof. a) Let X be a Noetherian topological space. We ﬁrst claim that X can
be expressed as a ﬁnite union of irreducible closed subsets. Indeed, consider the
collection of closed subsets of X which cannot be expressed as a ﬁnite union of
irreducible closed subsets. If this collection is nonempty, then by Proposition 297
there exists a minimal element Y . Certainly Y is not itself irreducible, so is the
union of two strictly smaller closed subsets Z1 and Z2 . But Z1 and Z2 , being
strictly smaller than Y , must therefore be expressible as ﬁnite unions of irreducible
closed subsets and therefore so also can Y be so expressed, contradiction.
b) So write
X = A1 ∪ . . . ∪ An
where each Ai is closed and irreducible. If for some i ̸= j we have Ai ⊂ Aj , then
we call Ai redundant and remove it from our list. After a ﬁnite number of such
removals, we may assume that the above ﬁnite covering of X by closed irreducibles is
irredundant in the sense that there are no containment relations between distinct
∪n
Ai ’s. Now let Z be any irreducible closed subset. Since Z = i=1 (Z ∩ Ai ) and Z is
irreducible, we must have Z = Z ∩ Ai for some i, i.e., Z ⊂ Ai . It follows that the
“irredundant” Ai ’s are precisely the maximal irreducible closed subsets, i.e., the
irreducible components.
We deduce the following important result, which is not so straightforward to prove
using purely algebraic methods:
Corollary 301. Let I be a proper ideal in a Noetherian ring R. The set of prime
ideals p which are minimal over I (i.e., minimal among all prime ideals containing
I ) is ﬁnite and nonempty. 184 PETE L. CLARK Exercise: Prove Corollary 301.
13.6. The map on spectra induced by a homomorphism.
13.7. Hochster’s Theorem.
A topological space X is sober if for every irreducible closed subspace Y of X ,
there exists a unique point y ∈ Y such that Y = {y }. Equivalently, a sober space
is one for which every irreducible closed subset has a unique generic point.
Exercise X.X:
a) Show that any Hausdorﬀ space is sober.
b) Show that a sobser space is Kolmogorov.
c) Show that the coﬁnite topology on an inﬁnite set is separated but not sober.
Exercise (Sobriﬁcation): For any topological space X , let X t be the set of irreducible closed subsets of X . There is a natural map t : X → X t via x → {x}.
Give X t the ﬁnal topology with respect to t, i.e., the ﬁnest topology that makes t
continuous. (Explicitly, a subset V of X t is open iﬀ its preimage in X is open.)
a) Show the map t induces a bijection from the open subsets of X to the open
subsets of X t .
b) Show that X t is a sober space.
c) Show that t is universal for continuous maps from X to a sober topological
space: i.e., for every sober space Y and continuous f : X → Y , there exists a
unique continuous F : X t → Y such that f = F ◦ t. Thus X t is (unfortunately!)
called the sobriﬁcation of X .
A topological space X is spectral if:
(SS1) X is quasicompact,
(SS2) X is sober, and
(SS3) The family of quasicompact open subsets of X is closed under ﬁnite intersections and is a base for the topology.
Remark: A Bourbakiste would insist that (SS3) =⇒ (SS1) by taking the empty
intersection. But we will not do so.
Exercise: Show that a ﬁnite space is spectral iﬀ it is T0 .
The following result gives an arguably cleaner characterization of spectral spaces.
Proposition 302. For a topological space X , TFAE:
(i) X is homeomorphic to an inverse limit of ﬁnite T0 spaces.
(ii) X is spectral.
Exercise X.X: Prove Proposition 302.48
Proposition 303. For any ring R, Spec R is spectral.
48I presume this is a relatively easy exercise working directly from the deﬁnitions, but I have
not attempted it myself. Please contact me if you have tried to prove this result but failed. COMMUTATIVE ALGEBRA 185 Exercise: Prove Proposition 303. (Hint: you will ﬁnd the needed results in the previous subsections. Especially, use Proposition 296 to prove the sobriety of Spec R.)
For any ring R we endow the set MaxSpec(R) of maximal ideals of R with the topology it inherits as a subset of Spec(R). When necessary, we describe MaxSpec R as
the “maximal spectrum” of R.
Proposition 304. For any ring R, MaxSpec R is separated and quasicompact.
Theorem 305. (Hochster’s Thesis [Ho69])
a) A spectral topological space is homeomorphic to the prime spectrum of some ring.
b) A separated quasicompact space is homeomorphic to the maximal spectrum of
some ring.
We do not aspire to give a proof of Theorem 305 at this time. (If you know of a
nice, short proof, please tell me!)
Exercise: Show that every compact (this includes Hausdorﬀ!) space X is homeomorphic to MaxSpec(C (X )), where C (X ) is the ring of continuous Rvalued functions on X .
Exercise:
a) Show that the specialization relation gives an equivalence of categories between
the category of T0 ﬁnite spaces and the category of ﬁnite partially ordered sets.
b)* Formulate a generalization of part a) in which T0 ﬁnite spaces are replaced
by T0 Alexandroﬀ spaces. (A topological space is Alexandroﬀ if an arbitrary
intersection of closed subsets is closed.)
Exercise: Let n ∈ Z+ .
a) Use Hochster’s Thesis and the previous exercise to show that there exists a ring
R with exactly n prime ideals p1 , . . . , pn such that p1 ⊂ p2 ⊂ . . . ⊂ pn .
b) For n = 1, 2, exhibit Noetherian rings with these properties. For n ≥ 3, show
that there is no such Noetherian ring.
13.8. The support of a module.
Let M be a module over a ring R. For m ∈ M , we put
ann m = {x ∈ R  xm = 0}.
Also we deﬁne
∩ ann M = {x ∈ R  xM = 0} so that ann M = m∈M ann m. (We have seen this concept before: we deﬁned a
module to be faithful if ann M = 0.)
Exercise X.X: ∩ {ωi }i∈I be a set of generators for M as an Rmodule. Show
Let
that ann M = i∈I ann ωi .
For a module M over a ring R, we deﬁne its support
supp M = {p ∈ Spec R  Mp ̸= 0}. 186 PETE L. CLARK This is an extremely important geometric notion which alas, gets very short shrift
in these notes. But we will establish the following basic result.
Proposition 306. For a ﬁnitely generated Rmodule M ,
supp M = V (ann M ).
In particular, supp M is a Zariski closed set.
Proof. Write M = ⟨ω1 , . . . , ωn ⟩R . For p ∈ Spec R, we have p ∈ supp M iﬀ Mp ̸= 0
iﬀ there exists i such that the image of ωi in Mp is not zero iﬀ there exists i such
that ann(ωi ) ⊂ p iﬀ
n
∩
ann M =
ann(ωi ) ⊂ p.
i=1 13.9. Rank functions revisited.
Theorem 307. Let M be a ﬁnitely generated module over a ring R.
a) For each n ∈ N, the set
Ur = {p ∈ Spec R  Mp can be generated over Rp by at most r elements}
is open in Spec R.
b) If M is ﬁnitely presented (e.g. if R is Noetherian), then the set
UF = {p ∈ Spec R  Mp is a free Rp module}
is open in Spec R.
Proof. (Matsumura) Suppose Mp = ⟨ω1 , . . . , ωr ⟩Rp . Each ωi is of the form mii with
s
×
mi ∈ M and si ∈ R \ p. But since si ∈ Rp for all i, we also have ⟨m1 , . . . , mr ⟩Rp =
Mp . Thus it is no loss of generality to assume that each ωi is the image in Mp of
an element of M . Let φ : Rr → M be the Rlinear map given by (a1 , . . . , ar ) →
∑
i ai ωi , and put C = coker φ, whence an exact sequence
Rr → M → C → 0.
Localizing this at a prime q of R gives an exact sequence
r
Rq → Mq → Cq → 0. When q = p we of course have Cq = 0. Moreover, C is a quotient of M hence a
ﬁnitely generated Rmodule, so by Proposition 306 its support supp C is a Zariskiclosed set. It follows that there exists an open neighborhood V of p such that
Cq = 0 for all q ∈ V .
b) Suppose that Mp is a free Rp module with basis ω1 , . . . , ωr . As above it is no
loss of generality to assume that each ωi is the image in Mp of an element of M .
Moreover, as we have also just seen, there exists a basic open neighborhood U (f )
such that for all q ∈ U (f ), the images of ω1 , . . . , ωr in Mq generate Mq as an Rq module. Replacing R by Rf and M by Mf we may assume that this occurs for all
q ∈ Spec R. Thus M/⟨ω1 , . . . , ωr ⟩R is everywhere locally zero, so it is locally zero:
M = ⟨ω1 , . . . , ωr ⟩. Deﬁning an Rlinear map φ : Rr → M as above and setting
K = Ker φ, we have the exact sequence
0 → K → Rr → M → 0. COMMUTATIVE ALGEBRA 187 Since M is ﬁnitely presented, according to Proposition 14 K is a ﬁnitely generated
Rmodule. Moreover we have Kp = 0 hence as above Kq = 0 for all q on some open
neighborhood V of p. By construction, for each q ∈ V , the images of ω1 , . . . , ωr in
Mq give an Rq basis for Mq .
Let M be a ﬁnitely generated, locally free module over a ring R. Earlier we deﬁned
the rank function r : Spec R → N as the rank of the free, ﬁnitely generated Rp module Mp . Applying Theorem 307a) to the locally free module M says precisely
that the rank function is lowersemicontinuous : it can jump up upon specialization,
but not jump down.
We now ask the reader to look back at Theorem 183 and see that for a ﬁnitely
generated module M over a general ring R, M is projective iﬀ it is locally free and
ﬁnitely presented. When R is Noetherian, being ﬁnitely presented is equivalent to
being ﬁnitely generated, so being projective is the same as being locally free. However, in the general case we have had little to say about the distinction between
ﬁnitely presented and ﬁnitely generated modules. Is there some way to rephrase
the subtly stronger property of ﬁnite presentation, perhaps a more geometric way?
Indeed there is:
Theorem 308. Let M be a ﬁnitely generated locally free Rmodule. TFAE:
(i) The rank function rM : Spec R → N is locally constant.
(ii) M is a projective module.
Proof. (i) =⇒ (ii): By Theorem 183, it is enough to show that for all m ∈
MaxSpec R, there exists f ∈ R \ m such that Mf is a free module. Let n = r(m),
and let x1 , . . . , xn be an Rm basis for Mm . Choose X1 , . . . , Xn ∈ M such that for
×
all i, the image of Xi in Mm is of the form ui xi for uu ∈ Rm . Let u : Rn → M
be the map sending the ith standard basis element ei to Xi . Since M is ﬁnitely
n
generated, by Proposition 181 there exists f ∈ R \ m such that uf : Rf → Mf
is surjective. It follows that for all g ∈ R \ m, uf g is surjective. Moreover, by
hypothesis there is some such g such that r(p) = n for all p ∈ X (g ). Replacing f
n
by f g we may assume that r(p) = n for all p ∈ X (f ). For all such p, up : Rp → Mp
is therefore a surjective endomorphism from a rank n free module to itself. Since
ﬁnitely generated modules are Hopﬁan, up is an isomorphism. By the local nature
of isomorphisms (Proposition 175) we conclude uf is an isomorphism, so Mf is free.
(ii) =⇒ (i): By Theorem 183, M is Zlocally free: there exists a ﬁnite Zfamily
{fi }∏ I such that for all i ∈ I , Mfi is ﬁnitely generated and free. Thus the modi∈
n
ule i=1 Mfi is ﬁnitely generated and projective over the faithfully ﬂat Ralgebra
∏n
i=1 Rfi , so by faithfully ﬂat descent (Theorem 109) M itself is projective.
Corollary 309. Let R be a ring with Spec R irreducible (e.g. a domain). For a
ﬁnitely generated Rmodule M , TFAE:
(i) R is projective.
(ii) R is Zlocally free.
(iii) R is locally free.
(iv) R is ﬂat.
Exercise: Prove Corollary 309. 188 PETE L. CLARK 14. Integrality in Ring Extensions
14.1. First properties of integral extensions.
If S is a ring extension of R – i.e., R ⊂ S – we will say that an element α of
S is integral over R if there exist a0 , . . . , an−1 ∈ R such that
αn + an−1 αn−1 + . . . + a1 α + a0 = 0.
Note that every element α ∈ R satisﬁes the monic polynomial t − α = 0, so is
integral over R.
Theorem 310. Let R ⊂ T be an inclusion of rings, and α ∈ T . TFAE:
(i) α is integral over R.
(ii) R[α] is ﬁnitely generated as an Rmodule.
(iii) There exists an intermediate ring R ⊂ S ⊂ T such that α ∈ S and S is ﬁnitely
generated as an Rmodule.
(iv) There exists a faithful R[α]submodule M of T which is ﬁnitely generated as
an Rmodule.
Proof. (i) =⇒ (ii): If α is integral over R, there exist a0 , . . . , an−1 ∈ R such that
αn + an−1 αn−1 + . . . + a1 α + a0 = 0,
or equivalently
αn = −an−1 αn−1 − . . . − a1 α − a0 .
This relation allows us to rewrite any element of R[α] as a polynomial of degree at
most n − 1, so that 1, α, . . . , αn−1 generates R[α] as an Rmodule.
(ii) =⇒ (iii): Take T = R[α].
(iii) =⇒ (iv): Take M = S .
(iv) =⇒ (i): Let m1 , . . . , mn be a ﬁnite set of generators for M over R, and express
each of the elements mi α in terms of these generators:
αmi = n
∑ rij mj , rij ∈ R. j =1 Let A be the n × n matrix αIn − (rij ); then recall from linear algebra that
AA∗ = det(A) · In ,
where A∗ is the “adjugate” matrix (of cofactors). If m = (m1 , . . . , mn ) (the row
vector), then the above equation implies 0 = mA = mAA∗ = m det(A) · In . The
latter matrix equation amounts to mi det(A) = 0 for all i. Thus • det(A) = •0 on
M , and by faithfulness this means det(A) = 0. Since so that α is a root of the
monic polynomial det(T · In − (aij )).
Exercise: Let S be a ﬁnitely generated Ralgebra. Show that TFAE:
(i) S/R is integral.
(ii) S is ﬁnite over R (as an Rmodule!).
In particular, if S/R is an extension and α1 , . . . , αn are all integral over R, then
R[α1 , . . . , αn ] is a ﬁnitely generated Rmodule. COMMUTATIVE ALGEBRA 189 Proposition 311. (Integrality is preserved under quotients and localizations)
Let S/R be an integral ring extension.
a) Let J be an ideal of S . Then S/J is an integral extension of R/(J ∩ R).
b) Let T be a multiplicatively closed subset of nonzero elements of R. Then ST is
an integral extension of RT .
Proof. a) First note that the kernel of the composite map R → S → S/J is J ∩ R,
so that R/(J ∩ R) → S/J is indeed a ring extension. Any element of S/J is of the
form x + J for x ∈ S , and if P (t)tn + an−1 tn−1 + . . . + a1 t + a0 = 0 ∈ R[t] is a
polynomial satisﬁed by x, then reducing coeﬃcientwise gives a monic polynomial
P (t) ∈ R/(J ∩ R) satisﬁed by x.
b) Let J = {s ∈ S ∃t ∈ T  ts = 0}, an ideal of S . Then ST is a quotient ring of
S/J and RT is a quotient ring of R/(J ∩ R), by part a) we are reduced to the case
in which R → RT and S → ST . Let x , x ∈ S, y ∈ T be an element of ST . Let
y
P (t) = tn + an−1 tn−1 + . . . + a0 ∈ R[t] be a monic polynomial satisﬁed by S . Then
( )n
( )n−1
x
an−1 x
a0
+
+ . . . + n = 0,
y
y
y
y
showing that x
y is integral over RT . Lemma 312. Let R ⊂ S ⊂ T be an inclusion of rings. If α ∈ T is integral over
R, then it is also integral over S .
Proof. If α is integral over R, there exists a monic polynomial P ∈ R[t] such that
P (α) = 0. But P is also a monic polynomial in S [t] such that P (α) = 0, so α is
also integral over S .
Lemma 313. Let R ⊂ S ⊂ T be rings. If S is ﬁnitely generated as an Rmodule
and T is ﬁnitely generated as an S module then T is ﬁnitely generated as an Rmodule.
Proof. If α1 , . . . , αr generates S as an Rmodule and β1 , . . . , βs generates T as an
S module, then {αi βj }{ 1 ≤ i ≤ r, 1 ≤ j ≤ s} generates T as an Rmodule: for
α ∈ T , we have
∑
∑∑
α=
bj βj =
(aij αi )βj ,
j i j with bj ∈ S and aij ∈ R.
Corollary 314. (Transitivity of integrality) If R ⊂ S ⊂ T are ring extensions such
that S/R and T /S are both integral, then T /R is integral.
Proof. For α ∈ S , let αn + bn−1 αn−1 + . . . + b1 α + b0 = 0 be an integral dependence relation, with bi ∈ S . Thus R[b1 , . . . , bn−1 , α] is ﬁnitely generated over
R[b1 , . . . , bn−1 ]. Since S/R is integral, R[b1 , . . . , bn−1 ] is ﬁnite over R. By Lemma
313, R[b1 , . . . , bn−1 , α] is a subring of T containing α and ﬁnitely generated over R,
so by Theorem 310, α is integral over R.
Corollary 315. If S/R is a ring extension, then the set IS (R) of elements of S
which are integral over R is a subring of S , the integral closure of R in S .
Thus R ⊂ IS (R) ⊂ S . 190 PETE L. CLARK Proof. If α ∈ S is integral over R, R[α1 ] is a ﬁnitely generated Rmodule. If α2 is
integral over R it is also integral over R[α1 ], so that R[α1 ][α2 ] is ﬁnitely generated
as an R[α1 ]module. By Lemmma 313, this implies that R[α1 , α2 ] is a ﬁnitely
generated Rmodule containing α1 ± α2 and α1 · α2 . By Theorem 310, this implies
that α1 ± α2 and α1 α2 are integral over R.
If R ⊂ S such that IS (R) = R, we say R is integrally closed in S .
Proposition 316. Let S be a ring. The operator R → IS (R) on subrings of R is
a closure operator in the abstract sense, namely it satisﬁes:
(CL1) R ⊂ IS (R),
(CL2) R1 ⊂ R2 =⇒ IS (R1 ) ⊂ IS (R2 ).
(CL3) IS (IS (R)) = IS (R).
Proof. (CL1) is the (trivial) Remark 1.1. (CL2) is obvious: evidently if R1 ⊂ R2 ,
then every element of S which satisﬁes a monic polynomial with R1 coeﬃcients
also satisﬁes a monic polynomial with R2 coeﬃcients. Finally, suppose that α ∈ S
is such that αn + an−1 αn−1 + . . . + a1 α + a0 = 0 for ai ∈ IS (R). Then each ai
is integral over R, so R[a1 , . . . , an ] is ﬁnitely generated as an Rmodule, and since
R[a1 , . . . , an , α] is ﬁnitely generated as an R[a1 , . . . , an ]module, applying Lemma
313 again, we deduce that α lies in the ﬁnitely generated Rmodule R[a1 , . . . , an , α]
and hence by Theorem 310 is integral over R.
14.2. Integral closure of domains.
Until further notice we restrict to the case in which R ⊂ S are integral domains.
Proposition 317. Let R ⊂ S be an integral extension of integral domains.
a) R is a ﬁeld iﬀ S is a ﬁeld.
b) An extension of ﬁelds is integral iﬀ it is algebraic.
Proof. a) Suppose ﬁrst that R is a ﬁeld, and let 0 ̸= α ∈ S . Since α is integral over
R, R[α] is ﬁnitely generated as an Rmodule, and it is wellknown in ﬁeld theory that
this implies R[α] = R(α). Indeed, taking the polynomial of least degree satisﬁed
by α, say α(αn−1 + an−1 αn−2 + . . . + a1 ) = −a0 , then 0 ̸= a0 ∈ R is invertible, so
−1 n−1
(α
+ an−1 αn−2 + . . . + a1 ) = α−1 ,
a0
and S is a ﬁeld. Conversely, if S is a ﬁeld and a ∈ R, then R[a−1 ] is ﬁnitedimensional over R, i.e., there exist ai ∈ R such that
a−n = an−1 a−n+1 + . . . + a1 a−1 + a0 .
Multiplying through by an−1 gives
a−1 = an−1 + an−2 a + . . . + a1 an−2 + a0 an−1 ∈ R,
completing the proof of part a). Over a ﬁeld every polynomial relation can be
rescaled to give a monic polynomial relation, whence part b).
Remark: A more sophisticated way of expressing Proposition 317 is that if S/R is
an integral extension of domains, then dim R = 0 iﬀ dim S = 0. Later we will see
that in fact dim R = dim S under the same hypotheses.
The integral closure of one ﬁeld R in another ﬁeld S is called the algebraic closure COMMUTATIVE ALGEBRA 191 of R in S .
Exercise:
a) Let S/R be an extension of ﬁelds. If S is algebraically closed, then so is IS (R).
b) Deduce that if R = Q, S = C, then IS (R) is an algebraically closed, algebraic
extension of Q, denoted Q and called the ﬁeld of all algebraic numbers.
Theorem 318. Let S/R be an extension of integral domains, and let T ⊂ R be
a multiplicatively closed subset. Then IT −1 S (T −1 R) = T −1 IS (R). In other words,
localization commutes with integral closure.
Proof. Let K be the fraction ﬁeld of R and L the fraction ﬁeld of S . Then T −1 IS (R)
is precisely the subring of L generated by T −1 and the elements of S which are integral over R. Since both of these kinds of elements of T −1S are integral over T −1 R
and integral elements form a subring, we must have T −1 IS (R) ⊂ IT −1 S (T −1 R).
Conversely, let s ∈ T −1 S be an element which is integral over T −1 R, so that there
t
exist a0 , . . . , an−1 ∈ T −1 R such that
s
s
( )n + an−1 ( )n−1 + . . . + a0 = 0.
t
t
Clearing denominators, we get
sn + (tan−1 )sn−1 + . . . + tn = 0,
which shows that s is integral over R, so s
t ∈ T −1 IS (R). Proposition 319. Suppose S/R is an extension with R a domain and S a ﬁeld.
Let K be the fraction ﬁeld of R and M the fraction ﬁeld of S . Then the fraction
ﬁeld of IS (R) is IM (K ).
Proof. Let us write L for the quotient ﬁeld of IS (R). First we show IM (K ) ⊂ L:
let 0 ̸= x ∈ IS (K ), so there exist a0 , . . . , an−1 ∈ K such that xn + an−1 xn−1 + . . . +
a1 x + a0 = 0. After clearing denominators and relabelling, we get a0 , . . . , an ∈ R
such that
an xn + an−1 xn−1 + . . . + a1 x + a0 = 0.
Multiplying through by an−1 , we get
n
(an x)n + an−1 (an x)n−1 + . . . + a1 an−2 (an x) + an−1 a0 = 0,
n
n
which shows that an x ∈ IS (R), so x ∈ IS (R) · (R \ 0)−1 hence is in the fraction ﬁeld
of IS (R).
The reverse inclusion L ⊂ IM (K ) is quite similar, since an arbitrary nonzero
element x of L is of the form α with α, β integral over R. But then certainly α and
β
β are both integral over K – i.e., algebraic over K , and then so also are β −1 and
x = α . So again x satisﬁes a polynomial an tn + . . . with coeﬃcients in R and then
β
an x is integral over R, hence an x and an are algebraic over K and thus x = an x is
x
algebraic over K so lies in IM (K ).
Example: If R = Z, S = C, then IS (R) is called the ring of all algebraic integers, and often denoted Z. By Proposition 319, its fraction ﬁeld is the ﬁeld Q of
all algebraic numbers. This turns out to be a very interesting ring, and it will crop
up several times in the sequel as an example or counterexample. For instance: 192 PETE L. CLARK Exercise 1.1.3: Show that Z is not ﬁnitely generated as a Zmodule.49
This shows that an integral extension need not be ﬁnite but only “locally ﬁnite”
(here we are using the word “locally” in a rough intuitive sense. not in the sense
of localization in commutative algebra).
Corollary 320. Let R be a domain with fraction ﬁeld K and M/K an arbitrary
ﬁeld extension. Put S = IM (R). Then S is integrally closed.
Proof. By Proposition 319, the ﬁeld of fractions L of S is the algebraic closure of
K in M . If x ∈ L is integral over R, then since L ⊂ M it lies in IM (R) = S .
Remark on terminology: It is unfortunate that the word “integral” in commutative algebra is used both to describe rings without zero divisors and also – quite
distinctly – to describe an extension satisfying a certain kind of ﬁniteness condition. The ﬁrst use of the word “integral” is essentially redundant in the algebraic
setting: if we just said “domain” instead of “integral domain”, the meaning would
be the same. However, in geometric language one has the notion of an “integral
scheme”; in the case of the aﬃne scheme Spec R associated to a commutative ring
R, the meaning of this is precisely that R be a domain, and because of this it is
undesirable to banish the “integral” from “integral domain.” Perhaps it would be
better instead to replace the term “integral extension.” We leave it as an exercise
to the reader to suggest some alternate terminology: this is less silly than it sounds,
because in order to do this one needs to grapple with the question, “What is an
integral extension, really?” It is perhaps telling that there is, so far as I know, no
notion of an “integral scheme extension” in algebraic geometry.
Let R be an integral domain with fraction ﬁeld K . We say that R is integrally
closed if IK (R) = R, i.e., if any element of the fraction ﬁeld satisfying a monic integral polynomial with Rcoeﬃcients already belongs to R. It follows immediately
from Proposition 316 that in any case IK (R) is integrally closed.
√
Exercise 1.1.4: Let R = Z[ −3] = Z[t]/(t2 + 3). Show that R is not integrally
closed, and compute its integral closure.
The geometric terminology for an integrally closed domain is normal. The process
of replacing R by its integral closure IK (R) is often called normalization.
14.3. Spectral properties of integral extensions.
Going down (GD): If we have I1 ⊃ I2 of R and J1 ∈ Spec S such that J1 ∩ R = I1 ,
there exists J2 ∈ Spec S such that J2 ⊂ J1 and J2 ∩ R = I2 .
Lemma 321. Let R be a local ring with maximal ideal p and S/R an integral
extension. Then the pushed forward ideal pS is proper.
∑
Proof. Suppose not: then there exist pi ∈ p, si ∈ S such that 1 =
i si pi .
Therefore any counterxample would take place already in the ﬁnite Rmodule
49In fact it is not even a Noetherian ring, so not even ﬁnitely generated as a Zalgebra. COMMUTATIVE ALGEBRA 193 R[s1 , . . . , sd ]. By induction on d, it is enough to consider the case of n = 1:
S = R[s]. Consider as usual a relation
(17) sn = an−1 sn−1 + . . . + a1 s + a0 , ai ∈ R of minimal possible degree n. If 1 ∈ pS then we have
(18) 1 = p0 + p1 s + . . . + pk sk , pi ∈ p. In view of (17) we may assume k ≤ n − 1. Since 1 − p0 is not in the maximal ideal
of the local ring R, it is therefore a unit; we may therefore divide (18) by 1 − p0
and get an equation of the form
1 = p′ s + . . . + p′ sq , p′ ∈ p.
1
q
i
This shows that s ∈ S × . Replacing a0 = a0 · 1 in (17) by a0 (p′ s + . . . + p′ sq ), we
q
1
get an integral dependence relation which is a polynomial in s with no constant
term. Since s is a unit, we may divide through by it and get an integral dependence
relation of smaller degree, contradiction.
Theorem 322. An integral ring extension S/R satisﬁes property (LO):50 every
prime ideal p of R is of the form S ∩ P for a prime ideal P of S .
Proof. For p a prime ideal of R, we denote – as usual – by Rp the localization of R
at the multiplicatively closed subset R \ p. Then Rp is local with unique maximal
ideal pRp , and if we can show that there exists a prime ideal Q of Sp lying over
pRp , then the pullback P = Q ∩ S to S is a prime ideal of S lying over p. By
Lemma 321, there exists a maximal ideal Q ⊃ pS and then Q ∩ R is a proper ideal
containing the maximal ideal p and therefore equal to it.
Corollary 323. (Going up theorem of CohenSeidenberg [CS46]) Let S/R be an
integral extension and p ⊂ q be two prime ideals of R. Let P be a prime ideal of S
lying over p (which necessarily exists by Theorem 322). Then there exists a prime
ideal Q of S containing P and lying over q.
Proof. Apply Theorem 322 with R = R/p S = S/P and p = q/p.
Corollary 324. (Incomparability) Suppose S/R is integral and P ⊂ Q are two
primes of S . Then P ∩ R ̸= Q ∩ R.
Proof. By passage to S/P , we may assume that P = 0 and S is an integral domain,
and our task is to show that any nonzero prime ideal P of S lies over a nonzero
ideal of R. Indeed, let 0 ̸= x ∈ P , and let P (t) = tn + an−1 tn−1 + . . . + a0 ∈ R[t]
be a monic polynomial satisﬁed by x; we may assume a0 ̸= 0 (otherwise divide by
t). Then a0 ∈ xS ∩ R ⊂ P ∩ R.
Corollary 325. Let S/R be an integral extension, P a prime ideal of S lying over
p. Then P is maximal iﬀ p is maximal.
Proof. First proof: Consider the integral extension S/P /(R/p); the assertion to be
proved is that S/P is a ﬁeld iﬀ R/p is a ﬁeld. But this is precisely Proposition 317a).
Second proof: If p is not maximal, it is properly contained in some maximal ideal
q. By the Going Up Theorem, there exists a prime Q ⊃ P lying over q, so P
is not maximal. Conversely, suppose that p is maximal but P is not, so there
50Or, lying over. 194 PETE L. CLARK exists Q
P . Then Q ∩ R is a proper ideal containing the maximal ideal p, so
Q ∩ R = p = P ∩ R, contradicting the Incomparability Theorem.
Invoking Going Up and Incomparability to (re)prove the elementary Corollary 325
is overkill, but these more sophisticated tools also prove the following
Corollary 326. Let S/R be an integral extension of rings. Then the Krull dimensions of R and S are equal.
Proof. Recall that the Krull dimension of a ring is the supremum of the length of a
ﬁnite chain of prime ideals. Suppose p0 p1 . . . pd are primes in R. Applying
Theorem 322, we get a prime P0 of S lying over p0 , and then repeated application
of the Going Up Theorem yields a chain of primes P0
P1
...
Pd , so that
dim(S ) ≥ dim(R). Similarly, if we have a chain of prime ideals P0
. . . Pd of
length d in S , then the Incomparability Theorem 324 implies that for all 0 ≤ i < d,
Pi ∩ R Pi+1 .
14.4. Integrally closed domains.
Let S/R be an extension of integral domains. Immediately from the deﬁnition
of integrality, there is a very concrete way to show that an element x ∈ S is integral
over R: we just have to exhibit a monic polynomial P ∈ R[t] such that P (x) = 0.
What if we want to show that an algebraic element x is not integral over R? It
would suﬃce to show that R[x] is not a ﬁnite Rmodule, but exactly how to do this
is not clear.
√
As an example, it is obvious that α = 2 is an algebraic integer, but unfortu√
nately it is not obvious that β = 22 is not an algebraic integer. (And of course we
√
need to be careful, because e.g. γ = 1+2 5 is an algebraic integer, since it satisﬁes
t2 + t − 1 = 0.) One thing to notice is that unlike α and γ , the minimal polynomial
of β , t2 − 1 , does not have Zcoeﬃcients. According to the next result, this is
2
enough to know that β is not integral over Z.
Theorem 327. Let R be a domain with fraction ﬁeld K , S/R an extension ring,
and x ∈ S an integral element over R.
a) Let P (t) ∈ K [t] be the minimal polynomial of x over K . Then P (t) ∈ IK (R)[t].
b) So if R is integrally closed, the minimal polynomial of x has Rcoeﬃcients.
Proof. a) We may assume without loss of generality that S = R[x]; then, by integrality, S is a ﬁnite Rmodule. Let L be the fraction ﬁeld of S , d = [L : K ], and
let s1 , . . . , sd be a K basis for L consisting of elements of S . Because the minimal
polynomial P (t) is irreducible over K , it is also the characteristic polynomial of
x· viewed as a K linear automorphism of L. Evidently the matrix Mx of x· with
respect to the basis s1 , . . . , sd has coeﬃcients in S ∩ K , hence in IK (R). The coefﬁcients of P (t) are polynomials in the entries of the matrix Mx , hence they also lie
in IK (R). Part b) follows immediately.
Exercise X.X.X: Let R be a domain with fraction ﬁeld K . Let S/R be an extension
such that for every x ∈ S which is integral over R, the minimal polynomial P (t) ∈
K [t] has Rcoeﬃcients. Show that R is integrally closed. COMMUTATIVE ALGEBRA 195 Theorem 328. (Local nature of integral closure) For an integral domain R, TFAE:
(i) R is integrally closed.
(ii) For all prime ideals p of R, Rp is integrally closed.
(iii) For all maximal ideals m of R, Rm is integrally closed.
Proof. Let K be the fraction ﬁeld of R. Assume (i), and let p ∈ Spec R. By
Theorem 318, the integral closure of Rp in K is Rp . Evidently (ii) =⇒ (iii).
Assume (iii), and let x be an element of K which is integral over R. Then for every
maximal ideal m of R, certainly x is integral over Rm , so by assumption x ∈ Rm
∩
∩
and thus x ∈ m Rm . By Corollary 176 we have m Rm = R.
Exercise X.X: Let R be an integrally closed domain with fraction ﬁeld K , L/K an
algebraic ﬁeld extension, S the integral closure of R in L and G = Aut(L/K ).
a) Show that for every σ ∈ G, σ (S ) = S .
b) For P ∈ Spec S and σ ∈ G, show σ (P ) = {σ (x)  x ∈ P} is a prime ideal of S .
c) Show that P ∩ R = σ (P ) ∩ R.
In conclusion, for every p ∈ Spec R, there is a welldeﬁned action of G on the
(nonempty!) set of prime ideals P of S lying over p.
Lemma 329. Let R be a domain with fraction ﬁeld K of characteristic p > 0, let
L/K be a purely inseparable algebraic extension of K (possibly of inﬁnite degree),
and let S be the integral closure of R in L. For any p ∈ Spec R, rad(pR) is the
unique prime of S lying over p.
Exercise: Prove Lemma 329. (Suggestions: recall that since L/K is purely insepa
arable, for every x ∈ L, there exists a ∈ N such that xp ∈ K . First observe
that rad(pR) contains every prime ideal of S which lies over p and then show that
rad(pR) is itself a prime ideal.)
Theorem 330. (Going Down Theorem of CohenSeidenberg [CS46]) Let R be an
integrally closed domain with fraction ﬁeld K , and let S be an integral extension of
R. If p1 ⊂ p2 are prime ideals of R and P2 is a prime ideal of S lying over p2 , then
there exists a prime ideal P1 of S which is contained in P2 and lies over p1 .
Proof. Let L be a normal extension of K containing S , and let T be the integral
closure of R in L. In particular T is integral over S , so we may choose Q2 ∈ Spec T
lying over P2 and also Q1 ∈ Spec T lying over p1 . By the Going Up Theorem
there exists Q′ ∈ Spec T containing Q1 and lying over p2 . Both Q2 and Q′ lie over
p2 , so by Theorem 342 there exists σ ∈ Aut(L/K ) such that σ (Q′ ) = Q2 . Thus
σ (Q1 ) ⊂ σ (Q′ ) = Q2 and σ (Q1 ) lies over p1 , so that setting P1 = σ (Q1 ) ∩ S we
have P1 ∩ R = p1 and P1 ⊂ σ (Q′ ) ∩ S = Q2 ∩ S = P2 .
Remark: In [?, Chapter 5] one ﬁnds a proof of Theorem 330 which avoids all Galoistheoretic considerations. However it is signiﬁcantly longer than the given proofs of
Theorems 342 and 330 combined and – to me at least – rather opaque.
14.5. The Noether Normalization Theorem.
14.5.1. The classic version.
Theorem 331. (Noether Normalization I) Let k be a ﬁeld, R = k [x1 , . . . , xm ] an
integral domain which is a ﬁnitely generated k algebra,51 and let K be the fraction
51Here the x ’s are not assumed to be independent indeterminates.
i 196 PETE L. CLARK ﬁeld of R.
a) There exists d ∈ Z, 0 ≤ d ≤ m, and algebraically independent elements y1 , . . . , yd ∈
R such that R is ﬁnitely generated as a module over the polynomial ring k [y1 , . . . , yd ]
– or equivalently, that R/k [y1 , . . . , yd ] is an integral extension.
b) The integer d is equal to both the Krull dimension of R and the transcendence
degree of K/k .
Proof. a) (Jacobson) The result is trivial if m = d, so we may suppose m > d.
Then the yi are algebraically dependent over k : there exists a nonzero polynomial
∑
f (s1 , . . . , sm ) =
aJ sj1 · · · sjm , aJ ∈ k [s1 , . . . , sm ]
m
1
with f (x1 , . . . , xm ) = 0. Let X be the set of monomials sJ = sj1 · · · sjm occuring
m
1
in f with nonzero coeﬃcients. To each such monomial we associate the univariate
polynomial
j1 + j2 t + . . . + jm tm−1 ∈ Z[t].
The polynomials obtained in this way from the elements of X are distinct. Since
a univariate polynomial over a ﬁeld has only ﬁnitely many zeroes, it follows that
there exists a ≥ 0 such that the integers j1 + j2 a + . . . + jm am−1 obtained from the
monomials in X are distinct. Now consider the polynomial
m− 1 f (s1 , sa + t1 , . . . , sa
1
1
We have
m−1 a
f (s1 , sa + t1 , . . . , s1
1 = ∑ + tm ) = ∑ + tm ) ∈ k [s, t].
m− 1 aJ sj1 (sd + t2 )j2 · · · (sa
1
1
1 + tm )jm J aJ sj1 +j2 a+...+jm a
1 m−1 + g (s1 , t2 , . . . , ym ), J m m−1
∑
in which the degree of g in s1 is less than that of J aJ sj1 +j2 a+...+j a
. Hence
1
m−1
+ tm ) is a monic polynomial in x1
for suitable β ∈ k × , βf (s1 , sa + t2 , . . . , sa
1
1
i− 1
with k [t2 , . . . , tm ]coeﬃcients. Putting wi = xi − xa
for 2 ≤ i ≤ m, we get
1
m− 1 βf (x1 , xd + w2 , . . . , xa
1
1 + wm ) = 0, so that x1 is integral over R′ = k [w2 , . . . , wm ]. By induction on the number of generators, R′ has a transcendence base {yi }d=1 such that R′ is integral over k [y1 , . . . , yd ].
i
Thus R is integral over k [y1 , . . . , yd ] by transitivity of integrality.
b) Since R/k [y1 , . . . , yd ] is an integral extension, by Corollary 326 the Krull dimension of R is equal to the Krull dimension of k [y1 , . . . , yd ], which by Corollary 285 is
d. Moreover, since R is ﬁnitely generated as a k [y1 , . . . , yd ], by Proposition 319 the
fraction ﬁeld K is ﬁnitely generated as a k (y1 , . . . , yd )module, so the transcendence
degree of K/k is equal to the transcendence degree of k (y1 , . . . , yd )/k , i.e., d.
14.5.2. Noether normalization over a domain.
Theorem 332. (Noether Normalization II) Let R ⊂ S be domains with S ﬁnitely
generated as an Ralgebra. There exists a ∈ R• and y1 , . . . , yd ∈ S algebraically
independent over the fraction ﬁeld of R such that Sa (the localization of S at the
multiplicative subset generated by a) is ﬁnitely generated as a module over T =
Ra [y1 , . . . , yd ]. COMMUTATIVE ALGEBRA 197 Proof. (K.M. Sampath) Let K be the fraction ﬁeld of R and let x1 , . . . , xm be a set
of Ralgebra generators for S . Then
S ′ := S ⊗R K = K [x1 , . . . , xm ]
is ﬁnitely generated over K (as above, the xi ’s need not be algebraically independent). Applying Theorem 331, we get algebraically independent elements y1 , . . . , yd ∈
S ′ such that S ′ is a ﬁnitely generated T ′ := K [y1 , . . . , yd ]module. Multiplying by
a suitable element of R× , we may assume yi ∈ S for all i.
Since S ′ is ﬁnitely generated as a T ′ module, it is integral over T ′ . For 1 ≤ i ≤ m,
xi satisﬁes a monic polynomial equation with coeﬃcients in T ′ :
n
n
y1 + Pi,1 (y1 , . . . , yd )yi −1 + . . . + Pi,n = 0. Let a be the product of the denominators of all coeﬃcients of all the polynomials
Pi,k . It follows that Sa is integral and ﬁnitely generated as a T = Ra [y1 , . . . , yd ]algebra, hence it is ﬁnitely generated as a T module.
Exercise: In the setting of Theorem 332 suppose that S is a graded Ralgebra.
Show that we may take all the yi to be homogeneous elements.
14.5.3. Applications. The Noether Normalization Theorem is one of the foundational results in algebraic geometry: geometrically, it says that every integral aﬃne
variety of dimension d is a ﬁnite covering of aﬃne dspace Ad . Thus it allows us to
study arbitrary varieties in terms of rational varieties via branched covering maps.
It is almost as important as a theorem of pure algebra, as even the “soft” part of
the result, that the Krull dimension of an integral aﬃne k algebra is equal to the
transcendence degree of its fraction ﬁeld, is basic and useful.
One of the traditional applications of Noether Normalization is to prove Hilbert’s
Nullstellensatz. As we have seen, it is fruitful to channel proofs of the Nullstellensatz through Zariski’s Lemma, and this is no exception.
Proposition 333. Noether Normalization implies Zariski’s Lemma.
Exercise: Prove Proposition 333.
Theorem 334. Let k be a ﬁeld, and let R be a domain which is ﬁnitely generated
as a k algebra, with fraction ﬁeld K . Then:
a) The Krull dimension of R is equal to the transcendence degree of K/k .
b) Every maximal chain of prime ideals in R has length dim R.
Proof. a) By Noether normalization, R is ﬁnite over k [t1 , . . . , td ], so K is ﬁnite
over k (t1 , . . . , td ). Thus the transcendence degree of K/k is d. On the other hand,
R is integral over k [t1 , . . . , td ] so by Theorem X.X dim R = dim k [t1 , . . . , td ]. By
Corollary XXX, dim k [t1 , . . . , td ] = d.
b) See MadapusiSampath, p. 119...
14.6. A Little Invariant Theory.
Let R be a commutative ring, let G be a ﬁnite group, and suppose G acts on
R by automorphisms, i.e., we have a homomorphism ρ : G → Aut(R). We deﬁne
RG = {x ∈ R ∀g ∈ G, gx = x}, 198 PETE L. CLARK the ring of Ginvariants – it is indeed a subring of R.
Remark: As in the case of rings acting on commutative groups, we say that Gaction on R is faithful if the induced homomorphism ρ : G → Aut(R) is injective.
Any Gaction induces a faithful action of G/ ker(ρ), so it is no real loss of generality to restrict to faithful Gactions. We will do so when convenient and in such a
situation identify G with its isomorphic image in Aut R.
The simplest case is that in which R = K is a ﬁeld. Then K G is again a ﬁeld
and K/K G is a ﬁnite Galois extension. Conversely, for any ﬁnite Galois extension
K/F , F = K Aut(K/F ) . This characterization of Galois extensions was used by
E. Artin as the foundation for an especially elegant development of Galois theory
(which swiftly became the standard one). Note also the analogy to topology: we
have the notion of a ﬁnite Galois covering Y → X of topological spaces as one for
which the group G = Aut(Y /X ) of deck transformations acts freely and properly
discontinuously on Y such that Y /G = X .
The branch of mathematics that deals with invariant rings under ﬁnite group actions is called classical invariant theory. Historically it was developed along
with basic commutative algebra and basic algebraic geometry in the early 20th
century, particularly by Hilbert. Especially, Hilbert’s work on the ﬁnite generation
of invariant rings was tied up with his work on the Basis Theorem.
For a ∈ R, deﬁne NG (a) =
have deﬁned a map ∏
σ ∈G σ (a). Observe that NG (a) ∈ RG , so that we NG : R → R G .
Note that NG is not a homomorphism of additive groups. However, when R is a
domain, there is an induced map
NG : R• → (RG )•
which is a homomorphism of monoids, so induces a homomorphism on unit groups.
Exercise X.X.: Let R[t] be the univariate polynomial ring over R. Show that there
is a unique action of G by automorphisms of G on R[t] extending the Gaction on
R and such that gt = t. Show that (R[t])G = RG [t].
Proposition 335. For a ﬁnite group G acting on R, R/RG is integral.
Proof. For x ∈ R, deﬁne
Φx (t) = NG (t − x) = ∏ (t − gx), g ∈G so Φx (t) ∈ (R[t])G = RG [t]. Thus Φx (t) is a monic polynomial with RG coeﬃcients
which is satisﬁed by x.
Base extension: Suppose that G is a ﬁnite group acting faithfully on R. Moreover,
let A be a ring and f : A → R be a ring homomorphism, so R is an Aalgebra.
Suppose moreover that f (A) ⊂ RG . In such a situation we say that G acts on R
by Aautomorphisms and write G ⊂ Aut(R/A). COMMUTATIVE ALGEBRA 199 Suppose we have another Aalgebra A′ . We can deﬁne an action of G on R ⊗A A′
by putting g (x ⊗ y ) := gx ⊗ y . We say that the Gaction is extended to A′ .
Proposition 336. In the above setup, suppose that A′ is a ﬂat Aalgebra. Then
there is a natural isomorphism
∼ RG ⊗A A′ → (R ⊗A A′ )G .
Proof. Madapusi p. 6566.
Corollary 337. Let G be a ﬁnite group acting on the ring R, and let S ⊂ RG be
a multiplicatively closed set. Then (S −1 R)G = S −1 RG .
Exercise: Prove Corollary 337.
In particular, suppose R is a domain with fraction ﬁeld K , and let F be the fraction
ﬁeld of RG . Then the Gaction on R extends to a Gaction on K , and Corollary
337 gives K G = F . Thus the invariant theory of integral domains is compatible
with the Galois theory of the fraction ﬁelds.
Proposition 338. If R is integrally closed, so is RG .
Proof. Let x ∈ K be an element which is integral over RG . Then x is certainly
also integral over R, and since R is integally closed in L we have x ∈ R. Thus
x ∈ R ∩ K = R ∩ LG = RG .
Theorem 339. (Noether [No26]) Suppose that R is a ﬁnitely generated algebra
over some ﬁeld k with k = k G . Then:
a) R is ﬁnitely generated as an RG module.
b) RG is a ﬁnitely generated k algebra.
Proof. a) Since R is a ﬁnitely generated k algebra and k ⊂ RG , R is a ﬁnitely
generated RG algebra. But by Proposition 335 R/RG is integral. So R is ﬁnitely
generated as an RG module.
b) By part a), the ArtinTate Lemma (Theorem 231) applies to the tower of rings
k ⊂ RG ⊂ R. The conclusion is as desired: RG is a ﬁnitely generated k algebra.
Remark: The title of [No26] mentions “characteristic p”. In fact, when k has characteristic 0 the result had been proven by Hilbert signiﬁcantly earlier [Hi90], and
moreover for certain actions of inﬁnite linear groups, like SLn (k ). But Noether’s
formulation and proof give an excellent illustration of the economy and power of
the commutative algebraic perspective.
Let us now make contact with the setup of “classical invariant theory”. Let k
be a ﬁeld, V a ﬁnitedimensional vector space and ρ : G → Autk (V ) a linear representation of G on V . Let k [V ] = Sym(V ∨ ) be the algebra of polynomial functions
on V . If we choose a k basis e1 , . . . , en of V and let x1 , . . . , xn be the dual basis
of V ∨ , then k [V ] = k [x1 , . . . , xn ] is a polynomial ring in n independent indeterminates. There is an induced action of G on k [V ], namely for f ∈ k [V ] we put
(gf )(x) = f (g −1 x).
All of our above results apply in this situation. Especially, Theorem 339 applies
to tell us that the ring k [V ]G is ﬁnitely generated as a k algebra, or a ﬁnite system of invariants. Of course, we did not so much as crease our sleeves, let alone 200 PETE L. CLARK roll them up, to establish this: for a concretely given ﬁnite group G and action
on a k vector space V , it is of interest to explicitly compute such a ﬁnite system.
Moreover, the polynomial ring k [V ] is integrally closed: in the next section we will
see that it is a unique factorization domain and that this is a stronger property.
Therefore Proposition 338 applies to show that k [V ]G is integrally closed. This is
actually quite a robust and useful procedure for producing integrally closed rings.
Example (the one example everyone knows): Let k be any ﬁeld, n ∈ Z+ , let V = k n ,
let G = Sn be the symmetric group, and let G act on V by permuting the standard
basis elements e1 , . . . , en . The induced action of G on k [V ] = k [x1 , . . . , xn ] is (up to
an inversion, or alternately a choice of left versus right action) the one obtained by
permuting the variables xi . An element of k [V ]G is called a symmetric polynomial or symmetric function. There are evident symmetric functions s1 , . . . , sn ,
∑
namely s1 (x) = x1 + . . . + xn , s2 (x) = i<j xi xj , and so forth: sk is a sum of
(n)
the k monomials of degree k . Then the fundamental theorem on symmetric
functions is that k [V ]G is a polynomial ring on the sk , i.e., k [V ]G = k [s1 , . . . , sn ],
and the sk ’s are independent indeterminates.52
The above example is wellknown and extremely useful, but gives a misleadingly
simple impression of classical invariant theory. One can ask how often the ring
of invariants of a ﬁnite group action on a polynomial ring is again a polynomial
ring, and there is a nice answer to this. But let’s back up a step and go back to
“rational invariant theory”: if G acts on k [x1 , . . . , xn ], then as above it also acts on
the fraction ﬁeld k (x1 , . . . , xn ) and we know that k (x1 , . . . , xn )/k (x1 , . . . , xn )G is a
ﬁnite Galois extension. But must k (x1 , . . . , xn )G itself be a rational function ﬁeld,
as it was in the example above? This is known as Noether’s Problem: it was
ﬁrst posed by E. Noether in 1913. It is natural and important, for an aﬃrmative
answer would allow us to realize every ﬁnite group as a Galois group (i.e., the automorphism group of a Galois extension) of Q thanks to a famous theorem of Hilbert.
For more than half of the twentieth century, Noether’s problem remained open.
Finally, in 1969 R.G. Swan (yes, the same Swan as before!) found a representation
of the cyclic group of order 47 on a ﬁnitedimensional Qvector space for which the
invariant ﬁeld is not a rational function ﬁeld [Sw69]. Too bad – this was arguably
the best shot that anyone has ever taken at the Inverse Galois Problem over Q.53
Example: Let k be a ﬁeld of characteristic diﬀerent from 2, let V = k 2 , and
consider the action of the twoelement group G = {±1} on V by −1 acting as the
scalar matrix −1. The induced action on k [V ] = k [x, y ] takes x → −x and y → −y .
This is, apparently, a not very interesting representation of a not very ineresting
group. But the invariant theory is very interesting!
Exercise: a) Show that k [V ]G is generated as a k algebra by x2 , y 2 and xy .
b) Show that k [V ]G is isomorphic to the k algebra k [A, B, C ]/(AB − C 2 ).
c) Show that nevertheless the fraction ﬁeld of k [V ]G is rational, i.e., is isomorphic
52This will not be proved here. Someday it will appear in my ﬁeld theory notes, but not today.
53Actually, Serre’s Topics in Galois Theory describes a conjecture of J.L. ColliotTh´l`ne –
ee roughly a weaker form of Noether’s problem – which would still imply that every ﬁnite group is
a Galois group over Q. I am not aware of any progress on this conjecture. COMMUTATIVE ALGEBRA 201 to k (X, Y ) for independent indeterminates X and Y .
Before signing oﬀ on our quick glimpse of classical invariant theory, we cannot
resist mentioning one more classic theorem in the subject. It answers the question:
when is the invariant subalgebra k [V ]G isomorphic to a polynomial algebra over k ?
Let ρ : G → GL(V ) be a faithful representation of G on a ﬁnitedimensional
k vector space V . An element g ∈ GL(V ) is a pseudoreﬂection if it has ﬁnite
order and pointwise ﬁxes a hyperplane W in V . (Equivalently, a pseudoreﬂection
has characteristic polynomial (t − 1)dim V −1 (t − ζ ), where ζ is a root of unity in k .)
Exercise X.X: If k = R, any nontrivial pseudoreﬂection has order 2 – i.e., it really is a hyperplane reﬂection.
A faithful representation ρ of G is a pseudoreﬂection representation of G if
ρ(G) is generated by pseudorﬂections.
Theorem 340. (ShephardToddChevalley) Let k be a ﬁeld, and let G be a ﬁnite
group with #G not divisible by the characteristic of k . Let ρ : G → GL(V ) be a
faithful representation of G on a ﬁnitedimensional vector space. TFAE:
(i) k [V ]G is a polynomial algebra.
(ii) k [V ] is a free k [V ]G module.
(iii) ρ is a pseudoreﬂection representation of G.
A proof in the case k = C is given in [Neus07, Chapter 13].
Exercise X.X: It follows from Theorem 340 the fundamental theorem on symmetric
functions that the standard permutation representation of the symmetric group Sn
on k n is a pseudoreﬂection representation. Show this directly.
14.7. Abstract algebraic number theory I: Galois extensions of integrally
closed domains.
Proposition 341. Let G be a ﬁnite group acting by automorphisms on a ring R,
with invariant subring RG . Let ι : RG → R, and let p ∈ Spec RG .
a) There is a natural action of G on the ﬁber (ι∗ )−1 (p) – i.e., on the set of primes
P of R such that ι∗ P = p.
b) The Gaction on the ﬁber (ι∗ )−1 (p) is transitive.
Proof. Let P ∈ Spec R and σ ∈ G. Deﬁne
σ P = {σx  x ∈ P}.
It is straightforward to verify that σ P is a prime ideal of R (if you like, this follows
from the fact that Spec is a functor). Moreover (σ P ) ∩ RG is the set of all elements
σx with x ∈ P such that for all g ∈ G, gσx = σx. As g runs through all elements
of G, so does gσ −1 , hence (σ P ) ∩ RG = P ∩ RG = p.
b) Let P1 , P2 be two primes of R lying over a prime p of RG . Let x ∈ P1 . Then
NG (x) ∈ P1 ∩ RG = p ⊂ P2 . Since P2 is prime, there exists at least one σ ∈ G
∪
such that σx ∈ P2 , and thus P1 ⊂ σ∈G σ P2 . By Prime Avoidance (Lemma 227),
there exists σ ∈ G such that P1 ⊂ σ P2 . Since R/RG is integral, Incomparability
(Corollary 324) yields P1 = σ P2 . 202 PETE L. CLARK Theorem 342. Let R be an integrally closed domain with fraction ﬁeld K , let L/K
be a normal algebraic ﬁeld extension (possibly of inﬁnite degree), and let S be the
integral closure of R in L. Let p ∈ Spec R, and let Xp be the set of all prime ideals
of S lying over p. Then G = Aut(L/K ) acts transitively on Sp .
Proof. Step 0: Do Exercise X.X to see that the action of G on Xp is welldeﬁned.
Step 1: Suppose [L : K ] = n < ∞, and write G = {σ1 = 1, . . . , σr }.54 Seeking a
−
contradiction, suppose there are P1 , P2 ∈ Xp such that P2 ̸= σj 1 P1 for all j . By
−1
Corollary 324, P2 is not contained in any σj P1 , so that by Lemma 227 (Prime
∪−
Avoidance) there exists x ∈ P2 \ j σj 1 P1 . Let q be the inseparable degree of
(∏
)q
L/K and put y =
j σj (x) . Thus y = NL/K (x), so y ∈ K . Moreover y is
integral over the integrally closed domain R, so y ∈ R. But since σ1 = 1, y ∈ P2 ,
so y ∈ P2 ∩ R = p ⊂ P1 , and thus, since P1 is prime, σj (x) ∈ P1 for some j :
contradiction!
Step 2: We will reduce to the case in which L/K is a Galois extension. Let G =
Aut(L/K ) and K ′ = LG , so that L/K ′ is Galois and K ′ /K is purely inseparable.
Let R′ be the integral closure of R in K ′ . Then by Lemma 329 Spec R′ → Spec R
is a bijection. So we may as well assume that K ′ = K and L/K is Galois.
Step 3: For each ﬁnite Galois subextension M of L/K , consider the subset
F (M ) := {σ ∈ G  σ (P1 ∩ M ) = P2 ∩ M }.
Observe that F (M ) is a union of cosets of Gal(L/M ) hence is (open and) closed
in the∏
Krull topology. By Step 1, we have F (M ) ̸= ∅. Moreover, the compositum
M
M = i Mi of any ﬁnite number {∩ i } of ﬁnite Galois subextenions is again a ﬁnite
Galois subextension, and we have i F (Mi ) ⊃ F (M ) ̸= ∅. Therefore as Mi ranges
through all ﬁnite Galois subextensions of L/K , {F (Mi )}i∈I is a family of closed
subsets of the compact space G satisfying the ﬁnite intersection condition, and it
∩
follows that there exists σ ∈ i F (Mi ) = F (L) i.e., σ ∈ G such that σ P1 = P2 .
15. Factorization
Let R be an integral domain, and x a nonzero, nonunit element of R. We say that
x is irreducible if for any y, z ∈ R such that x = yz , one of y or z is a unit.
Remarks: For any unit u ∈ R× , we get factorizations of the form x = u · (u−1 x), so
every nonzero nonunit has at least these factorizations, which we wish to regard as
“trivial.” On the other hand, y and z cannot both be units, for then x would also
be a unit.
Let us deﬁne a factorization of a nonzero nonunit a ∈ R as a product
a = x1 · · · xn ,
such that each xi is irreducible. We say that two factorizations
a = x1 · · · xn = y1 · · · ym
are equivalent if the multisets of associated principal ideals {{(xi )}} = {{(yj )}}
are equal. More concretely, this means that m = n and that there is a bijection
54Recall that we are assuming that L/K is normal, so L/K is separable iﬀ L/K is Galois iﬀ
r = n. COMMUTATIVE ALGEBRA 203 σ : {1, . . . , n} → {1, . . . , m} such that (yσ(i) ) = (xi ) for all 1 ≤ i ≤ n.
If factorizations always exist and any two factorizations of a given element are
equivalent, we say R is a unique factorization domain (UFD).
15.1. Kaplansky’s Theorem (II).
A basic and important result that ought to get covered at the undergraduate level
is that PID implies UFD. In fact this is easy to prove. What is more diﬃcult is to
get a sense of exactly how UFDs are a more general class of rings than PIDs. In
this regard, an elegant theorem of Kaplansky seems enlightening.
Exercise X.X: Let x be an element of an integral domain which can be expressed as
x = p1 · · · pn ,
such that for 1 ≤ i ≤ n, Pi = (pi ) is a prime ideal. If then there exist principal
prime ideals Q1 , . . . , Qm such that
(x) = Q1 · · · Qm ,
then m = n and there exists a permutation σ of the integers from 1 to n such that
Qi = Pσ(i) for all i.
Exercise X.X: Let R be a domain, and let S be the set of all nonzero elements
x in R such that (x) can be expressed as a product of principal prime ideals. Show
that S is a saturated multiplicatively closed subset.
Theorem 343. (Kaplansky) An integral domain is a UFD iﬀ every nonzero prime
ideal in R contains a prime element.
Proof. Suppose R is a UFD and 0 ̸= P is a prime ideal. Let x ∈ P be a nonzero
nonunit. Write
x = p1 · · · pr
a product of prime elements. Then x ∈ P implies pi ∈ P for some i, so (pi ) ⊂ P .
Conversely, assume each nonzero prime ideal of R contains a principal prime.
Let S be the set of all products of prime elements, so that by Exercise X.X, S is
a saturated multiplicative subset. By Exercise X.X, it is enough to show that S
contains all nonzero nonunits of R. Suppose for a contradiction that there exists a
nonzero nonunit x ∈ R \ S . The saturation of S implies S ∩ (x) = ∅, and then by
Theorem 117 there exists a prime ideal P containing x and disjoint from S . But
by hypothesis, P contains a prime element p, contradicting its disjointness from
S.
We deduce immediately:
Corollary 344. Let R be a domain.
a) If every ideal of R is principal, R is a UFD.
b) Conversely, if R is a UFD of dimension one, every ideal of R is principal.
Proof. a) Recall that a height one prime is a prime ideal p of R which does not
strictly contain any nonzero prime ideal. Applying Kaplansky’s theorem, we get a
prime element x such that 0 (x) ⊂ p, and height one forces equality. For part b),
it clearly suﬃces to show that a Noetherian domain in which each height one prime 204 PETE L. CLARK is principal is a UFD. But let p be any nonzero prime ideal; since R is Noetherian,
(DCC) holds on prime ideals, so p contains a prime ideal which has height one,
hence contains a prime element and we are done by Kaplansky’s theorem.
Remark: Applying b) to R = k [x1 , . . . , xn ], we get the important geometric fact
that each irreducible codimension one subvariety in aﬃne space is cut out by a
single equation.
15.2. Atomic domains, (ACCP). If every nonzero nonunit admits at least one
factorization, we say R is an atomic domain55.
Exercise X.X: Show that the ring Z of all algebraic integers is not an atomic domain. Indeed, since for every algebraic integer x, there exists an algebraic integer
y such that y 2 = x, there are no irreducible elements in Z!
The condition of factorization into irreducibles (in at least one way) holds in every
Noetherian domain. In fact, a much weaker condition than Noetherianity suﬃces:
Proposition 345. Let R be a domain in which every ascending chain of principal
ideals stabilizes. Then every nonzero nonunit factors into a product of irreducible
elements. In particular, a Noetherian domain is atomic.
Proof. Let R be a domain satisfying the ascending chain condition for principal
ideals (ACCP for short), and suppose for a contradiction that R is not an atomic
domain. Then the set of principal ideals generated by unfactorable elements is
nonempty, so by our assumption there exists a maximal such element, say I = (a).
Evidently a is not irreducible, so we can begin to factor a: a = xy where x and
y are nonunits. But this means precisely that both principal ideals (x) and (y )
properly contain (a), so that by the assumed maximality of (a), we can factor both
x and y into irreducibles: x = x1 · · · xm , y = y1 · · · yn . But then
a = x1 · · · xm y1 · · · yn
is a factorization of a, contradiction.
This proposition motivates us to consider also the class of domains which satisfy
the ascending chain condition for principal ideals (ACCP).
Exercise X.X: Suppose that R → S is an exension of rings such that S × ∩ R = R× .
(In particular, this holds for integral extensions.) Show that S satisﬁes (ACCP)
implies R satisﬁes (ACCP). Does the converse hold?
We have just seen that (ACCP) implies atomicity. The proof shows that under
(ACCP) we can always obtain an expression of a given nonzero nonunit by a ﬁnite
sequence of “binary factorizations” i.e., replacing an element x with y1 · y2 , where
y1 and y2 are nonunits whose product is x. After a bit of thought, one is inclined to
worry that it may be possible that this factorization procedure fails but nevertheless
irreducible factorizations exist. This worry turns out to be justiﬁed:
Theorem 346. There exists an atomic domain which does not satisfy (ACCP).
55In the ﬁrst pass on these notes, I used the sensible name “factorization domain.” But
apparently atomic domain is what is used in the literature, and since we ﬁnd this terminology
unobjectionable, we might as well use it here. COMMUTATIVE ALGEBRA 205 Proof. See [Gr74].
However, the following strenghtening of atomicity does imply ACCP:
A domain R is a bounded factorization domain BFD if it is atomic and for
each nonzero nonunit a ∈ R, there exists a positive integer N (a) such that in any
irreducible factorization a = x1 · · · xr we have r ≤ N (a).
Proposition 347. A UFD is a BFD.
Proof. An immediate consequence of the deﬁnitions.
Proposition 348. A BFD satisﬁes (ACCP).
Proof. Let R be a BFD. Suppose for a contradiction that (xi )i∈Z+ is a strictly
ascending chain of principal ideals. We therefore have
x0 = y1 x1 = y1 y2 x2 = . . . = y1 · · · yn xn = · · · ,
with each xi , yi a nonunit. Since R is atomic, we can reﬁne each factorization into
an irreducible factorization, but clearly an irreducible reﬁnement of y1 · · · yn xn has
at least n + 1 irreducible factors, contradicting BFD.
15.3. ELdomains.
An element x of a domain R is prime if the principal ideal (x) is a prime ideal.
Equivalently, x satisﬁes Euclid’s Lemma: if x  yz , then x  y or x  z .
Proposition 349. A prime element is irreducible.
Proof. If x is reducible, then x = yz with neither y nor z a unit, so that yz ∈ (x)
but y ̸ ∈(x), z ̸ ∈(x).
However, it need not be the case that irreducible elements are prime!
√
√
Example: Let R = Z[ −5]. Then the elements 2, 3 and 1 ± −5 are all irreducible, but the equation
√
√
2 · 3 = (1 + −5)(1 − −5)
shows that none of them are prime elements.
√
Exercise X.X: Check √ these assertions. Hint: Deﬁne N (a + b √−5) = a2 + 5b2 .
all
√
√
Check that N ((a + b −5)(c + d −5)) = N (a + b −5)N (c + d −5). Show that
√
α  β (in R) =⇒ N (α)  N (β ) (in Z), and use this to show that 2, 3, 1 ± −5 are
irreducible but not prime.
It is tempting to call a domain in which all irreducible elements are prime “Euclidean”, but this terminology is already taken for domains satisfying a generalization of the Euclidean algorithm (see XX). So we will, provisionally, call a ring in
which irreducible elements are prime an ELdomain. (EL = Euclid’s Lemma).
Theorem 350. For an integral domain R, TFAE:
(i) R is a UFD.
(ii) R satisﬁes (ACCP) and is an ELdomain.
(iii) R is an atomic ELdomain. 206 PETE L. CLARK Proof. i) =⇒ (ii): In the previous section we saw UFD =⇒ BFD =⇒ (ACCP).
We show UFD implies ELdomain: let x ∈ R be irreducible and suppose x  yz .
Let y = y1 · · · ym and z = z1 · · · zn be irreducible factorizations of y and z . Then
the uniqueness of irreducible factorization means that x must be associate to some
yi or to some zj , and hence x  y or x  z : R is an ELdomain.
(ii) =⇒ (iii) follows immediately from Proposition 345.
(iii) =⇒ (i): This is nothing else than the usual deduction of the fundamental
theorem of arithmetic from Euclid’s Lemma: in a factorization domain we have at
least one irreducible factorization of a given nonzero nonunit x. If we moreover
assume that irreducibles are prime, this allows us to compare any two irreducible
factorizations: suppose
x = y1 · · · ym = z1 · · · zn .
Then y1 is a prime element so divides zj for some j . WLOG, relabel to assume
j = 1. Since z1 is irreducible, we have y1 = u1 z1 and thus we may cancel to get
y2 · · · ym = (u−1 z2 )z3 · · · zn .
1
Continuing in this way we ﬁnd that each yi is associate to some zj ; when we get
∏
down to 1 = j zj we must have no factors of zj left, so m = n and R is a UFD.
15.4. GCDdomains.
For elements a and b of R, a greatest common divisor is an element d of R
such that: d  a, d  b and for any e ∈ R such that e  a, e  b, then e  d.
Exercise X.X: Show that if there exists a greatest common divisor of two elements
a and b, then an element d′ of R is a gcd of a and b iﬀ (d) = (d′ ). In particular,
any two gcd’s are associate.
If a and b have a gcd, it would be more logically sound to write gcd(a, b) to mean
the unique principal ideal whose generators are the various gcd’s of a and b. It is
traditional however to use the notation gcd(a, b) to denote an element, with the
understanding that in general it is only welldeﬁned up to multiplication by a unit.
(In some rings, principal ideals have canonical generators: e.g. in the integers we
may take the unique positive generator and in k [t] we may take the unique monic
generator. Under these circumstances, one generally means gcd(a, b) to stand for
this canonical generator.)
More generally, for elements a1 , . . . , an in a domain R, a greatest common divisor is an element d of R such that d  ai for all i and if e  ai for all i then e  d.
If a GCD of (a1 , . . . , an ) exists, it is unique up to associates, and we denote it by
gcd(a1 , . . . , an ). As above, it can be characterized as the unique minimal principal
ideal containing ⟨a1 , . . . , an . Moreover, these setwise GCDs can be reduced to pairwise GCD’s, e.g.:
Exercise: Let a, b, c be elements of a domain R and assume that all pairwise GCD’s
exist in R. Then gcd(a, b, c) exists and we have gcd(a, gcd(b, c)) = gcd(a, b, c) =
gcd(gcd(a, b), c).
A domain R is a GCDdomain if for all a, b ∈ R, gcd(a, b) exists. By the above COMMUTATIVE ALGEBRA 207 remarks, it would be equivalent to require that gcd(a1 , . . . , an ) for all ntuples of
elements in R.
Proposition 351. (GCD identities) Let R be a GCDdomain. Then:
a) For all a, b, c ∈ R, gcd(ab, ac) = a gcd(b, c).
a
b
b) For all a, b ∈ R \ {0}, gcd( gcd(a,b) , gcd(a,b) ) = 1.
c) For all a, b, c ∈ R, gcd(a, b) = gcd(a, c) = 1, then gcd(a, bc) = 1.
d) For all a, b, c ∈ R, gcd(a, b + ac) = gcd(a, b).
e) For all a, a1 , . . . , an , b1 , . . . , bn , c ∈ R, gcd(a, b1 +ca1 , . . . , bn +can ) = gcd(a, b1 , . . . , bn ).
Proof. a) Let x = gcd(ab, ac). Then a  ab and a  ac so a x: say ay = x. Since
x  ab and x  ac, y  b and y  c, so y  gcd(b, c). If z  b and z  c, then az  ab and
1
az  ac, so az  x = ay and z  y . Therefore gcd(b, c) = y = a gcd(ab, ac).
Part b) follows immediately.
As for part c): suppose gcd(a, b) = gcd(a, c) = 1, and let t divide a and bc. Then t
divides ab and bc so t  gcd(ab, bc) = b gcd(a, c) = b. So t divides gcd(a, b) = 1.
d) If d divides both a and b, it divides both a and b + ac. If d divides both a and
b + ac, it divides b + ac − c(a) = b.
e) We have
gcd(a, b1 + ca1 , . . . , bn + can ) = gcd(a, gcd(a, b1 + ac1 ), . . . , gcd(a, bn + acn ))
= gcd(a, gcd(a, b1 ), . . . , gcd(a, bn )) = gcd(a, b1 , . . . , bn ). Proposition 352. A GCDdomain is an ELdomain.
Proof. This follows from the fact gcd(x, y ) = gcd(x, z ) = 1 =⇒ gcd(x, yz ) = 1.
Theorem 353. Consider the following conditions on a domain R:
(i) R is a UFD.
(ii) R is a GCDdomain.
(iii) R is an ELdomain: irreducible elements are prime.
Then (i) =⇒ (ii) =⇒ (iii) always; and (iii) =⇒ (i) if R satisﬁes (ACCP).
In particular, a Noetherian domain is a UFD iﬀ it is GCD iﬀ it is EL.
Proof. Assume (i), and let x and y be nonzero elements of R. We may write
x = f1 · · · fr g1 · · · gs , y = uf1 · · · fr h1 · · · ht ,
where the f ’s, g ’s and h’s are prime elements, (gj ) ̸= (hk ) for all j, k and u ∈ R× .
It is then easy to check that f1 · · · fr is a gcd for x and y . (ii) implies (iii) is
Proposition 352. (iii) + (ACCP) implies (i) is Theorem 350.
We can now present some simple results that are long overdue. An extremely useful
fact in algebra is that any UFD is integrally closed in its fraction ﬁeld. We give a
slightly stronger result and then recall a classical application:
Theorem 354. A GCDdomain is integrally closed.
Proof. Let R be a gcddomain with fraction ﬁeld K . Suppose x ∈ K satisﬁes an
equation
xn + an−1 xn−1 + . . . + a1 x + a0 = 0, ai ∈ R. 208 PETE L. CLARK We may write x = s with s, t ∈ R, t ̸= 0. Proposition XXa) asserts that, after
t
dividing through by the gcd we may assume gcd(s, t) = 1. Substituting s for x and
t
clearing denominators as usual, we get
(
)
sn = − an−1 tsn−1 + . . . + a1 tn−1 s + a0 tn ,
and we have the usual contradiction: t divides the right hand side, so t divides sn .
But by Proposition XXc) gcd(s, t) = 1 implies gcd(sn , t) = 1. The only way out of
a contradiction is that t is a unit, i.e., x ∈ R.
Corollary 355. An algebraic integer which is a rational number is an integer:
Z ∩ Q = Z.
Exercise X.X: Prove Corollary 355.
√
Thus e.g. one can derive the irrationality of 2: it is a root of the monic polynomial
equation t2 − 2 = 0 but evidently not an integer, so cannot be rational.
Proposition 356. (Compatibility of GCD’s with localization) Let R be a GCDdomain and S a multiplicative subset of R. Then:
a) The localization S −1 R is again a GCDdomain.
b) For all x, y ∈ R• , if d is a GCD for x and y in R, then it is also a GCD for x
and y in S −1 R.
Exercise X.X: Prove Proposition 356.
15.5. Polynomial rings over UFDs.
Our goal in this section to show that if R is a UFD, then a polynomial ring in
any number (possibly inﬁnite) of indeterminates is again a UFD. This result generalizes the “undergraddy” fact that if k is a ﬁeld, k [t] is a UFD; the corresponding
fact that polynomials in k [t1 , . . . , tn ] factor uniquely into irreducibles is equally
basic and important, and arguably underemphasized at the pregraduate level (including high school, where factorizations of polynomials in at least two variables
certainly do arise).
If we can establish that R a UFD implies R[t] a UFD, then an evident induction argument using the “identity” R[t1 , . . . , tn , tn+1 ] = R[t1 , . . . , tn ][tn+1 ] gives us
the result for polynomials in ﬁnitely many indeterminates over a UFD (which is, so
far as I know, the most important case). It is then straightforward to deduce the
case for arbitrary sets of indeterminates.
There are several diﬀerent ways to prove the univariate case. Probably the most
famous is via Gauss’ Lemma. To give this argument, we need some preliminary
terminology.
Let R be any integral domain, and consider a nonzero polynomial
f = an tn + . . . + a1 t + a0 ∈ R[t].
We say f is primitive if x ∈ R, x  ai for all i implies x ∈ R× . In a GCDdomain, this is equivalent to gcd(a1 , . . . , an ) = 1. In a PID, this is equivalent
to ⟨a0 , . . . , an ⟩ = R. For a general domain, this latter condition is considerably
stronger: e.g. the polynomial xt + y ∈ k [x, y ][t] is primitive but the coeﬃcients do COMMUTATIVE ALGEBRA 209 not generate the unit ideal. Let us call this latter – usually too strong condition –
naively primitive.
Proposition 357. Let R be a domain, and f, g ∈ R[t] be naively primitive. Then
f g is naively primitive.
Proof. Suppose f and g are naively primitive but f g is not. Then by deﬁnition the
ideal generated by the coeﬃicents of f is proper, so lies in some maximal ideal m
of R. For g ∈ R[t], write g for its image in the quotient ring (R/m)[t]. Then our
assumptions give precisely that f , g ̸= 0 but f g = f g = 0. Thus f and g are zero
divisors in the integral domain (R/m)[t], a contradiction.
If R is a GCDdomain and 0 ̸= f ∈ R[t], we can deﬁne the content c(f ) of f to be
the gcd of the coeﬃcients of f , welldetermined up to a unit. Thus a polynomial is
primitive iﬀ c(f ) = 1.
Exercise X.X: Let R be a GCDdomain and 0 ̸= f ∈ R[t].
a) Show that f factors as c(f )f1 , where f1 is primitive.
b) Let 0 ̸= a ∈ R. Show that c(af ) = ac(f ).
Theorem 358. (Gauss’ Lemma) Let R be a GCDdomain. If f, g ∈ R[t] are
nonzero polynomials, we have c(f g ) = c(f )c(g ).
If we assume the stronger hypothesis that R is a UFD, we can give a very transparent proof along the lines of that of Proposition 357 above. Since this special
case may be suﬃcient for the needs of many readers, we will give this simpler proof
ﬁrst, followed by the proof in the general case.
Proof. (Classical proof for UFDs) The factorization f = c(f )f1 of Exercise X.X
reduces us to the following special case: if f and g are primitive, then so is f g .
Suppose that f g is not primitive, i.e., there exists a nonzero nonunit x which divides
all of the coeﬃcients of f g . Since R is a UFD, we may choose a prime element π  x.
Now we may argue exactly as in the proof of Proposition 357: (R/(π )[t] is an integral
domain, f and g are nonzero, but f g = f g = 0, a contradiction.
And now the proof in the general case, which makes use of some GCD identities.
Proof. (Haible) As above, we may assume that f = an tn + . . . + a1 t + a0 , g =
bm tm + . . . + b1 t + b0 ∈ R[t] are both primitive, and we wish to show that f g =
cm+n tm+n + . . . + c1 t + c0 is primitive. We go by induction on n. Since a primitive
polynomial of degree 0 is simply a unit in R, the cases m = 0 and n = 0 are both
trivial; therefore the base case m + n = 0 is doubly so. So assume m, n > 0. By
Proposition X.X(x), we have
c(f g ) = gcd(cn+m , . . . , c0 ) =
gcd(an bm , gcd(cn+m−1 , . . . , c0 ))  gcd(an , gcd(cn+m−1 , . . . , c0 ))·gcd(bm , gcd(cn+m−1 , . . . , c0 )).
Now
gcd(an , gcd(cn+m−1 , . . . , c0 )) = gcd(an , cn+m−1 , . . . , c0 )
= gcd(an , cn+m−1 − an bm−1 , . . . , cn − an b0 , cn−1 , . . . , c0 )
= gcd(an , c((f − an tn )g ).
Our induction hypothesis gives c((f − an tn )g ) = c(f − an tn )c(g ) = c(f − an tn ), so
gcd(an , cn+m−1 −an bm−1 , . . . , cn −an b0 , cn−1 , . . . , c0 ) = gcd(an , c(f −an tn )) = c(f ) = 1. 210 PETE L. CLARK Similarly we have gcd(bm , gcd(cn+m−1 , . . . , c0 )) = 1, so c(f g ) = 1.
Corollary 359. Let R be a GCDdomain with fraction ﬁeld K , and let f ∈ R[t]
be a polynomial of positive degree.
a) The following are equivalent:
(i) f is irreducible in R[t].
(ii) f is primitive and irreducible in K [t].
b) The following are equivalent:
(i) f is reducible in K [t].
(ii) There exist g, h ∈ R[t] such that deg(g ), deg(h) < deg(f ) and f = gh.
Proof. a) Assume (i). Clearly an imprimitive polynomial in R[t] would be reducible
in R[t], so f irreducible implies c(f ) = 1. Suppose f factors nontrivially in K [t], as
f = gh, where both g, h ∈ K [t] and have smaller degree than f . By Exercise X.X, we
may write g = c(g )g1 , h = c(h)h1 , with g1 , h1 primitive, and then f = c(g )c(h)g1 h1 .
But then g1 , h1 , being primitive, lie in R[t], and c(f ) = c(g )c(h) = c(gh) ∈ R, so
the factorization takes place over R[t], contradiction. (ii) =⇒ (i) is similar but
much simpler and left to the reader.
b) That (ii) =⇒ (i) is obvious, so assume (i). Because we can factor out the
content, it is no loss of generality to assume that f is primitive. Let f = g1 h1 with
g1 , h1 ∈ K [t] and deg(g1 ), deg(h1 ) < deg(f ). Because R is a GCDdomain, we may
˜
g
h
write g = d˜ , h = d2 with g , h ∈ R[t] primitive. Then we have d1 d2 f = g h, and
˜˜
˜˜
1
×
equating contents gives (d1 d2 ) = (1), so d1 , d2 ∈ R and thus the factorization
f = gh has the properties we seek.
Theorem 360. (Gauss) If R is a UFD, so is R[t].
We will give three diﬀerent proofs of this important result, essentially due to Gauss.
The ﬁrst proof is via Gauss’s Lemma.
Proof. Let K be the fraction ﬁeld of R, and let f ∈ R[t]• . We know that K [t] is a
PID hence a UFD, so we get a factorization
f = cg1 · · · gr ,
with c ∈ R and each gi ∈ R[t] is primitive and irreducible. Then factoring c into
irreducibles gives an irreducible factorization of f . If we had another irreducible
factorization f = dh1 · · · hs , then unique factorization in K [t] gives that we have
r = s and after permuting the factors have gi = ui hi for all i, where ui ∈ K × .
Since both gi and hi are primitive, we must have ui ∈ R× , whence the uniqueness
of the factorization.
We note that this proof relies on knowing that K [t] is a UFD, which of course
follows from the fact that polynomial division gives a Euclidean algorithm, as one
learns in an undergraduate course. This is of course an adaptation of the proof that
the ring Z is a UFD (the Fundamental Theorem of Arithmetic) essentially due
to Euclid.
But it is of interest to ﬁnd alternate routes to basic and important results.
Theorem 361. Let R be a domain with fraction ﬁeld K .
a) If R is an ACCPdomain, so is R[t].
b) If R is a GCDdomain, so is R[t].
c) Thus, once again, if R is a UFD, so is R[t]. COMMUTATIVE ALGEBRA 211 Proof. a) In an inﬁnite ascending chain {(Pi )} of principal ideals of R[t], deg Pi is
a descending chain of nonnegative integers, hence eventually constant. Therefore
for suﬃciently large n we have Pn = an Pn+1 with an ∈ R and (an+1 ) ⊃ (an ). Since
R is an ACCP domain, we have (an ) = (an+1 ) for suﬃciently large n, hence also
(Pn ) = (Pn+1 ) for suﬃciently large n.
b) (Haible, [Hai94]) Let f, g ∈ R[t]. We may assume that f g ̸= 0. As usual, write
˜
˜
f = c(f )f and g = c(g )˜. Since K [t] is a PID, may take the gcd of f and g in
g
˜
×
˜
˜
K [t], say d. The choice of d is unique only up to an element of K , so by choosing
˜
the unit appropriately we may assume that d lies in R[t] and is primitive. We put
˜.
d = gcd(c(f ), c(g )))d
˜
˜
˜
Step 1: We claim that d is a gcd of f and g in R[t]. Since d  f in K [t], we
˜
˜
f
˜
˜
may write d = a q with a, b ∈ R \ {0} and q ∈ R[t] primitive. Since bf = ad, we
˜
b
b
×
˜
˜
˜˜
have (b) = c(bf ) = c(ad) = (a), i.e., a ∈ R and thus d  f in R[t]. Similarly
˜  g . Moreover, sice d ∈ f K [t] + g K [t], there exist u, v ∈ R[t] and c ∈ R \ {0}
˜
˜
d˜
˜
˜
˜
˜
˜
with cd = uf + v g . Suppose h ∈ R[t] divides both f and g . Then h  cd, and
˜
˜
˜) = (1). Writing cd = a q with q ∈ R[t] primitive, and equating contents
c(h)  c(f
h
b
˜
d
˜
˜
in bcd = ahq , we get (bc) = (a), hence h = ab q ∈ R[t], so h  d.
c
Step 2: We claim that d is a gcd of f and g in R[t]. Certainly we have
˜
˜
(d) = (gcd(c(f ), c(g )d)  (c(f )f ) = (f ),
˜
so d  f . Similarly d  g . Conversely, let h ∈ R[t] divide f and g . Write h = c(h)h
˜ ∈ R[t] primitive. From h  f it follows that c(h)  c(f ) and thus h  f .
˜˜
for h
˜˜
˜˜
Similarly h  f so h  g . Thus c(h)  gcd(c(f ), c(g )), h  d and thus ﬁnally h  d.
c) If R is a GCD domain and an ACCP domain, it is also an atomic EL domain,
hence a UFD by Theorem 350.
Lindemann [Li33] and Zermelo [Ze34] (independently) gave (similar) striking proofs
of the Fundamental Theorem of Arithmetic avoiding all lemmas and packaging the
Euclidean division into a single inductive argument. Later several authors have
recorded analogous proofs of Gauss’s Theorem (Theorem 360): the earliest instance
we are aware of in the literature is due to S. Borofsky [Bor50]. We give a third,
“lemmaless” proof of Theorem 360 here.
Proof. It suﬃces to show that R[t] is an ACCP domain and an ELdomain. By
Theorem 361a), R[t] is an ACCP domain. Now, seeking a contradiction, we suppose
that R is an ELdomain but R[t] is not. Among the set of all elements in R[t]
admitting inequivalent irreducible factorizations, let p be one of minimal degree.
We may assume
p = f1 · · · fr = g1 · · · gs ,
where for all i, j , (fi ) ̸= (gj ) and
m = deg f1 ≥ deg f2 ≥ . . . ≥ deg fr ,
n = deg g1 ≥ deg g2 ≥ . . . ≥ deg gs ,
with n ≥ m > 0. Suppose the leading coeﬃcient of f1 (resp. g1 ) is a (resp. b). Put
q = ap−bf1 xn−m g2 · · · gs = f1 (af2 · · · fr −bxn−m g2 · · · gs ) = (ag1 −bf1 xn−m )g2 · · · gs .
Thus q = 0 implies ag1 = bf1 xn−m . If, however, q ̸= 0, then
deg(ag1 − bf1 xn−m ) < deg g1 , 212 PETE L. CLARK hence deg q < deg p and q has a unique factorization into irreducibles, certainly
including g2 , · · · , gs and f1 . But then f1 must be a factor of ag1 − bf1 xn−m and
thus also of ag1 . Either way ag1 = f1 h for some h ∈ R[t]. Since a is constant and f1
is irreducible, this implies h = ah2 , so ag1 = f1 ah2 , or g1 = f1 h2 , contradiction.
Corollary 362. Let R be a UFD and let {ti }i∈I be any set of indeterminates.
Then S = R[{ti }] is a UFD.
Proof. The case in which I is ﬁnite follows ∪
immediately from Theorem XX by induction. When I is inﬁnite, we have S = J R[{tj }], as J ranges over all ﬁnite
subsets of I : this is just because any given polynomial can only involve ﬁnitely
many indeterminates. For f ∈ S , let J be such that f ∈ R[{tj }], so f = p1 · · · pr is
a factorization into prime elements of R[{tj }]. Any two factorizations in S would
themselves lie in some subalgebra involving ﬁnitely many determinates, so the factorization must be unique.
Remark: In particular, a polynomial ring in inﬁnitely many variables over a ﬁeld
is a standard example of a nonNoetherian UFD.
We want to mention without proof two negative results.
Theorem 363. a) (Roitman) There exists an integrally closed atomic domain R
such that R[t] is not atomic.
b) (AndersonQuinteroZafrullah) There exists an EL domain R such that R[t] is
not an EL domain.
15.6. Application: Eisenstein’s Criterion.
The most famous criterion for irreducibility of univariate polynomials is named
after Ferdinand Eisenstein, who used it in an 1850 Crelle paper. In fact, the version for polynomials over Z was apparently ﬁrst given by T. Sch¨nemann in 1846
o
(also in Crelle ). Nowadays it is common to state and prove a version of Eisenstein’s
criterion with respect to a prime ideal in a UFD. We give a slight generalization:
Theorem 364. (Sch¨nemannEisenstein Criterion) Let R be a domain with fraco
tion ﬁeld K , and let f (t) = ad td + . . . + a1 t + a0 ∈ R[t]. Suppose that there exists
a prime ideal p of R such that ad ̸∈ p, ai ∈ p for all 0 ≤ i < d and a0 ̸∈ p2 .
a) If f is primitive, then f is irreducible over R[t].
b) If R is a GCDdomain, then f is irreducible over K [t].
Proof. a) Suppose to the contrary that f is primitive and reducible over R[t]: i.e.,
there exists a factorization f = gh with g (t) = bm tm + . . . + b1 t + b0 , h(t) =
cn tm + . . . + c1 t + c0 , deg(g ), deg(h) < deg(f ) and bm cn ̸= 0. Since a0 = b0 c0 ∈ p \ p2 ,
it follows that exactly one of b0 , c0 lies in p: say it is c0 and not b0 . Moreover, since
ad = bm cn ̸∈ p, cn ̸∈ p. Let k be the least index such that ck ̸∈ p, so 0 < k ≤ n.
Then b0 ck = ak − (b1 ck−1 + . . . + bk c0 ) ∈ p. Since p is prime, it follows that at least
one of b0 , ck lies in p, a contradiction.
b) Suppose R is a GCDdomain and (seeking a contradiction) that f is reducible
over K [t]. By Corollary 359b), we may write f = gh with g, h ∈ R[t] and
deg(g ), deg(h) < deg(f ). Then the proof of part a) goes gives a contradiction.
Corollary 365. Let R be a GCDdomain containing a prime element π (e.g. a
UFD that is not a ﬁeld). Then the fraction ﬁeld K of R is not separably closed. COMMUTATIVE ALGEBRA 213 Proof. To say that π is a prime element is to say that the principal ideal p = (π ) is
a nonzero prime ideal. Then π ̸∈ p2 , so for all n > 1, Pn (t) = tn − π is Eisenstein
with respect to p and hence irreducible in K [t]. Choosing n to be prime to the
characteristic of K yields a degree n separable ﬁeld extension Ln := K [t]/(Pn ).
15.7. Application: Determination of Spec R[t] for a PID R.
Let R be a PID. We wish to determine all prime ideals of the ring R[t]. Let us begin
with some general structural considerations. First, R is a onedimensional Noetherian UFD; so by Theorems 218, 361 and 229, R[t] is a twodimensional Noetherian
UFD. Being a UFD, its height one ideals are all principal. Since it has dimension
two, every nonprincipal prime ideal is maximal. Therefore it comes down to ﬁnding
all the maximal ideals.
However, to avoid assuming the Dimension Theorem, we begin with milder hypotheses on a prime ideal P , following [R, pp. 2223].
Namely, let P be a nonzero prime ideal of R[t]. We assume – only! – that P
is not principal. By Theorem 343, we are entitled to a prime element f1 of P . Since
P = (f1 ), let f2 ∈ P \ (f1 ). Then gcd(f1 , f2 ) = 1: since gcd(f1 , f2 )  f1 , the only
̸
other possibility is (gcd(f1 , f2 )) = (f1 ), so f1  f2 and f2 ∈ (f1 ), contradiction.
First Claim: Let K be the fraction ﬁeld of R. The elements f1 and f2 are
also relatively prime in the GCDdomain K [t]. Indeed, suppose that f1 = hg1 ,
f2 = hg2 with h, g1 , g2 ∈ K [t] and h a nonunit. By Gauss’ Lemma, we may write
h = ah0 , g1 = b1 γ1 , g2 = b2 γ2 with a1 , b1 , b2 ∈ K and h0 , γ1 , γ2 primitive elements of R[t]. Again by Gauss’ Lemma, h0 γ1 and h0 γ2 are also primitive, so
f1 = hg1 = (ab1 )(h0 γ1 ) ∈ R[t], which implies that ab1 ∈ R. Similarly, ab2 ∈ R, so
h0 is a nonunit of R[t] which divides both f1 and f2 , contradiction.
Let M := ⟨f1 , f2 ⟩, and put m = M∩ R. It remains to show that, as the notation
suggests, M is a maximal ideal of R[t] and m is a maximal ideal of R.
Second Claim: m ̸= 0. Since K [t] is a PID and f1 , f2 are relatively prime in
K [t], there exist a, b ∈ K [t] such that af1 + bf2 = 1. Let 0 ̸= c ∈ R be an element which is divisible by the denominator of each coeﬃcient of a and b: then
(ca)f1 + (cb)f2 = c with ca, cb ∈ R, so that c ∈ m.
Now put p = P ∩ R, so p = (p) is a prime ideal of the PID R. Moreover,
p=P ∩R⊃M∩R=m 0, so p is maximal. Since P ⊃ p, P corresponds to a prime ideal in R[t]/pR[t] =
(R/p)[t], a PID. Therefore P is generated by p ∈ p and an element f ∈ R[t] whose
image in (R/p)[t] is irreducible.
We therefore have proved:
Theorem 366. Let R be a PID, and P ∈ Spec R[t]. Then exactly one of the
following holds:
(0) P has height 0: P = (0).
(i) P has height one: P = (f ), for a prime element f ∈ R[t]. 214 PETE L. CLARK (ii) P has height two: P = ⟨p, f ⟩, where p is a prime element of R and f ∈ R[t] is
an element whose image in (R/p)[t] is irreducible. Moreover both P and p := P ∩ R
are maximal, and [R[t]/P : R/p] < ∞.
Exercise: Suppose R has only ﬁnitely prime ideals, so is not a HilbertJacobson
ring. By Theorem 287, there is a maximal ideal m of R[t] such that m ∩ R = (0).
Find one, and explain where m ﬁts in to the classiﬁcation of Theorem 366.
15.8. Characterization of UFDs among Noetherian domains.
Theorem 367. For a Noetherian domain R, TFAE:
(i) Every height one prime ideal of R is principal.
(ii) R is a UFD.
Proof. . . .
15.9. Power series rings over UFDs.
Exercise X.X: Show that if R is ACCP, so is R[[t]].
In particular, if R is a UFD, R[[t]] is ACCP. But of course the more interesting question is the following: Must R[[t]] be a UFD?
In contrast to Gauss’s Theorem, whether a formal power series ring over a UFD
must be a UFD was a perplexing problem to 20th century algebraists and remained
open for many years. Some special cases were known relatively early on.
Theorem 368. (R¨ckert [R¨33], Krull [Kr37]) Let k be a ﬁeld, and let n be a
u
u
positive integer. Then k [[t1 , . . . , tn ]] is a UFD.
A signiﬁcant generalization was proved by Buchsbaum and Samuel, independently,
in 1961. We deﬁne the height of a prime ideal p in a ring R to be the supremum
of all nonnegative integers n such that there exists a strictly ascending chain of
prime ideals p0 p1 . . . pn = p. In a domain R, a prime ideal has height 0 iﬀ
it is the zero ideal. A Noetherian domain R is regular if for every maximal ideal
m of R, the height of m is equal to the dimension of m/m2 as a vector space over
the ﬁeld R/m.
Theorem 369. ([Buc61], [Sa61]) If R is a regular UFD, then so is R[[t]].
In the same 1961 paper, Samuel also provided the ﬁrst example of a UFD R for
which R[[t]] is not a UFD [Sa61, §4].
One may also ask about formal power series in countably inﬁnitely many indeterminates. Let k be a ﬁeld. There is more than one reasonable way to deﬁne such
a domain. One the one hand, one could simply take the “union” (formally, direct
limit) of ﬁnite formal power series rings k [[t1 , . . . , tn ]] under the evident inclusion
maps. In any element of this ring, only ﬁnitely many indeterminates appear. However, it is useful also to consider a larger ring, in which the elements are inﬁnite
formal k linear combinations of monomials ti1 · · · tin . Let us call this latter domain
k [[t1 , . . . , tn , . . .]].
In fact, we have seen this domain before: it is isomorphic to the Dirichlet ring Dk
of functions f : Z+ → k with pointwise addition and convolution product. To see
this, we use unique factorization in Z! Namely, write enumerate the prime numbers COMMUTATIVE ALGEBRA 215 ∏∞
{pi }∞ and write n ∈ Z+ as n = i=1 pai , where ai ∈ Z+ and ai = 0 for all
i=1
i
suﬃciently large i. Then the map which sends f ∈ Dk to the formal power series
∑
∏∞ ai
n∈Z+ f (n)
i=1 ti gives an isomorphism from Dk to k [[t1 , . . . , tn , . . .]]. In 1959,
E.D. Cashwell and C.J. Everett used Theorem 368 to prove the following result. A
key part of their proof was later simpliﬁed by C.F. Martin, who pointed out the
applicability of K¨nig’s Inﬁnity Lemma.
o
Theorem 370. ([CaEv59], [Mar71])
a) For any ﬁeld k , the ring of formal power series k [[t1 , . . . , tn , . . .]] is a UFD.
b) In particular, the ring DC = {f : Z+ → C} of airthmetic functions is a UFD.
In almost any ﬁrst number theory course one studies unique factorization and also
arithmetic functions, including the Dirichlet ring structure (which e.g. leads to an
immediate proof of the M¨bius Inversion Formula). That arithmetic functions are
o
themselves an example of unique factorization is however a very striking result that
does not seem to be wellknown to most students or practitioners of number theory.
I must confess, however, that as a working number theorist I know of no particular
application of Theorem 370. I would be interested to learn of one!
15.10. Nagata’s Criterion for UFDs.
Proposition 371. Let R be a domain, S a saturated multiplicative subset, and
f ∈ R \ S . If f is prime as an element of R, it is also prime as an element of RS .
Proof. Since f ∈ R \ S , f is not a unit in RS . Let α, β ∈ RS be such that f  αβ
1
2
3
in RS . So there exists γ ∈ RS such that γf = αβ ; putting α = x1 , β = x2 , γ = x3
s
s
s
and clearing denominators, we get s1 s2 x3 f = s3 x1 x2 , so f  r3 x1 x2 . If f  s3 , then
since S is saturated, f ∈ S , contradiction. So, being prime, f divides x1 or x2 in
1
2
R, hence a fortiori in RS and therefore it also divides either x1 or x2 in RS , since
s
s
these are associates to x1 and x2 .
Theorem 372. Every localization of a UFD is again a UFD.
Exercise: Prove Theorem 372. (Suggestions: one gets an easy proof by combining
Theorem 343 with Proposition 371. But the result is also rather straightforward to
prove directly.)
A saturated multiplicative subset S of R is primal56 if it is generated by the
units of R and by the prime elements of S .
Lemma 373. An irreducible element of a primal subset is prime.
Proof. Suppose S is primal and f ∈ S is irreducible. By deﬁnition, there exists a
unit u and prime elements π1 , . . . , πn such that f = uπ1 · · · πn . Since uπ1 is also
prime, we may as well assume that u = 1. Then, since f is irreducible, we must
have n = 1 and f = π1 .
Theorem 374. For an atomic domain R, the following are equivalent:
(i) Every saturated multiplicative subset of R is primal.
(ii) R is a UFD.
56This terminology is my invention: do you like it? 216 PETE L. CLARK Proof. Since the set R× of units is trivially generated by the empty set of prime
elements, both conditions hold if R is a ﬁeld, so let us now assume otherwise.
Assume (i). Then, since R is a factorization domain which is not a ﬁeld, there
exists an irreducible element f of R. Let S be the saturated multiplicative subset
generated by S , which consists of all units of R together with all divisors of positive
powers f n of f . Since S is primal and strictly contains R× , there must exist a
prime element π which divides f n for some n. In other words, f n ∈ πR, and since
πR is prime, we must have that f = xπ for some x ∈ R. Since f is irreducible we
must have x ∈ R× , i.e., f ∼ π and is therefore a prime element. So R is an ACCP
domain and an ELdomain and hence a factorization domain by Theorem ??.
Assume (ii), let S be a saturated multiplicative subset of R, and suppose that
a
a
f ∈ S \ R× . Then f = uπ1 1 · · · πnn where the πi ’s are prime elements. Since each
πi  f , πi ∈ S for all i. It follows that indeed S is generated by its prime elements
together with the units of R.
Because of Theorem 374, it is no loss of generality to restate Theorem 372 as: the
localization of a UFD at a primal subset is again a UFD. The following elegant
result of Nagata may be viewed as a converse.
Theorem 375. (Nagata [Nag57]) Let R be a factorization domain and S ⊂ R a
primal subset. If the localized domain RS is a UFD, then so is R.
Proof. By Theorem 350 it suﬃces to show that if f ∈ R is irreducible, f is prime.
Case 1: f ̸∈ S , so f is not a unit in RS . Since RS is a UFD, it is enough to show
1
2
that f is irreducible in RS . So assume not: f = x1 · x2 with x1 , x2 ∈ R \ S and
s
s
s1 , s2 ∈ S . hen s1 s2 f = x1 x2 . By assumption, we may write s1 = up1 · · · pm and
s2 = vq1 · · · qn , where u, v ∈ R× and pi , qj are all prime elements of R. So p1  x1 x2 ;
1
2
since p1 is a prime, we must have either x1 ∈ R or x2 ∈ R. Similarly for all the
p
q
other pi ’s and qj ’s, so that we can at each stage divide either the ﬁrst or the second
factor on the right hand side by each prime element on the left hand side, without
1
12
leaving the ring R. Therefore we may write f = ( uv ) x1 x2 where t1 , t2 are each
tt
products of the primes pi and qj , hence elements of S , and also such that t1  x1 ,
t2  x2 , i.e., the factorization takes place in R. Moreover, since xi ∈ R \ S and
ti ∈ S , xii is not even a unit in RS , hence a fortiori not a unit in R. Therefore we
t
have exhibited a nontrivial factorization of f in R, contradiction.
Case 2: f ∈ S . Since S is primal, by Lemma 373, f is prime.
Remark: If S is the saturation of a ﬁnitely generated multiplicative set, the hypothesis that R is a factorization domain can be omitted.
Application: Let A be a UFD and consider the polynomial ring R = A[t]. Put
S = A \ {0}. As for any multiplicative subset of a UFD, S is generated by prime
elements. But moreover, since A[t]/(πA[t]) ∼ (A/πA)[t], every prime element π of
=
A remains prime in A[t], so viewing S as the multiplicative subset of A[t] consisting
of nonzero constant polynomials, it too is generated by prime elements. But if F
is the fraction ﬁeld of A, RS = (A[t])S = F [t] which is a PID and hence a UFD.
Nagata’s theorem applied to R and S now tells us – for the third time! – that
R = A[t] is a UFD.
Nagata used Theorem 375 to study the coordinate rings of aﬃne quadric cones. COMMUTATIVE ALGEBRA 217 Let k be a ﬁeld of characteristic diﬀerent from 2, and let f (x) = f (x1 , . . . , xn ) ∈
k [x1 , . . . , xn ] be a quadratic form, i.e., a homogeneous polynomial of degree
2 with k coeﬃcients. We assume that f the associated bilinear form (x, y ) →
1
2 (f (x + y ) − f (x) − f (y )) is nonsingular. Equivalently, by making an invertible linear change of variables every quadratic form can be diagonalized, and a quadratic
form is nonsingular iﬀ it admits a diagonalization
(19) f (x) = a1 x2 + . . . + an x2 witha1 , . . . , an ∈ k × .
1
n We wish to study the aﬃne quadric cone associated to f , namely Rf = k [x]/(f ).
Note that if quadratic forms f and g are isometric – i.e., diﬀer by an invertible
linear change of variables – then Rf ∼ Rg , so we assume if we like that f is in
=
diagonal form as in (19) above. If n ≥ 3 then every nonsingular diagonal quadratic
polynomial is irreducible, so Rf is a domain. If k is quadratically closed – i.e.,
admits no proper quadatic extension – then conversely any binary (n = 2) quadratic
form is reducible, so Rf is not a domain. (If f is not quadratically closed, there
exist irreducible binary quadratic forms, but we will not consider them here.)
Theorem 376. Let f = f (x1 , . . . , xn ) ∈ C[x1 , . . . , xn ] be a nonsingular quadratic
form. Then Rf = k [x]/(f ) is a UFD iﬀ n ≥ 5.
Proof. By the remarks above, Rf is a domain iﬀ n ≥ 3, so we may certainly restrict
to this case. Because C is algebraically closed, every quadratic form in n ≥ 2
variables is isotropic, i.e., there exists 0 ̸= a ∈ k n such that f (a) = 0: indeed, the
ﬁrst n − 1 coordinates of a may be chosen arbitrarily. By an elementary theorem
in the algebraic theory of quadratic forms [Lam06, Thm. I.3.4], we may make a
change of variables to bring f into the form:
f (x) = x1 x2 + g (x3 , . . . , xn ).
Case 1: Suppose n = 3, so that
2
f (x) = x1 x2 − ax3 for some a ∈ k × . In this case, to show that Rf is not a UFD it suﬃces to
show that the images x1 , x2 , x3 of x1 , x2 , x3 in Rf are nonassociate irreducibles,
for then x1 x2 = ax3 2 exhibits a nonunique factorization! To establish this, regard
k [x1 , x2 , x3 ] as a graded Calgebra in the usual way – with x1 , x2 , x3 each of degree
1 – so that the quotient Rf by the homogeneous ideal (f ) inherits a grading. Since
x1 has degree 1, if it were reducible, it would factor as the product of a degree one
element c1 x1 + c2 x2 + x3 x3 + (f ) and a degree zero element r + (f ), and thus
(rc1 − 1)x1 + rc2 x2 + rc3 x3 ∈ (f ).
But the left hand side has degree 1, whereas all nonzero elements in (f ) have degree
2 or higher, so r ∈ C[x]× and therefore the factorization is trivial. The irreducibility
of x2 and x3 is proved in the same way. If x1 ∼ x3 in Rf , then we may divide both
sides of x1 x2 − ax3 2 by x1 and deduce that also x2 ∼ x3 . But in the quotient ring
Rf /(x3 ), x3 maps to 0 and x1 and x2 do not, contradiction. So Rf is not a UFD.
Case 2: Suppose n = 4, so f (x) = x1 x2 + g (x3 , x4 ), where g (x3 , x4 ) is a nonsingular
binary form. Here for the ﬁrst time we use the full strength of the quadratic closure
of k : since k × = k ×2 , any two nonsingular quadratic forms in the same number of
variables are isometric, so we may assume WLOG that
f (x) = x1 x2 − x3 x4 . 218 PETE L. CLARK Now we argue exactly as in Case 1 above: in Rf , the images x1 , x2 , x3 , x4 are all
nonassociate irredcuble elements, so x1 x2 = x3 x4 is a nonunique factorization.
Case 3: n ≥ 5. Then n − 2 ≥ 3, so g is irreducible in the UFD C[x3 , . . . , xn ], hence
also in C[x2 , x3 , . . . , xn ]. Therefore Rf /(x1 ) = C[x1 , . . . , xn ]/(x1 , f ) = C[x2 , . . . , xn ]/(g )
is a domain, i.e., x1 is a prime element. Moreover,
R[x1 −1 ] = C[x1 , . . . , xn , x−1 ]/(x1 x2 − g )
1
∼ C[x1 , . . . , xn , x−1 ]/(x2 − g ) ∼ C[x1 , x3 , . . . , xn , x−1 ]
=
=
1
1
x1
is a localization of the UFD C[x1 , x3 , . . . , xn ] hence a UFD. By Nagata’s Criterion
(Theorem 375), Rf itself is a UFD.
Now let k be an arbitrary ﬁeld of characteristic not 2 and f ∈ k [x1 , . . . , xn ] a
nonsingular quadratic form. Without changing the isomorphism class of Rq we may
diagonalize f ; moreover without changing the ideal (f ) we may scale by any element
of k × , so without loss of generality we need only consider forms x2 +a2 x2 +. . .+an x2 .
n
1
2
Theorem 377. Let k be a ﬁeld of characteristic diﬀerent from 2 and f = x2 +
1
a2 x2 + . . . + an x2 a nonsingular quadratic form over k . Put Rf = k [x]/(f ).
n
2
a) If n ≤ 2 then Rf is not an integrally closed domain.
b) If n = 3, Rf is a UFD iﬀ f is anistropic: ∀a ∈ k n , f (a) = 0 =⇒ a = 0.
c) (i) Suppose f = x2 − ax2 − bx2 − cx2 . If a is a square in k , then Rf is a UFD
4
3
2
1
iﬀ −bc is not a square in k .
(i) If none of a, b, c, −ab, −ac, −bc is a square in k , then Rf is a UFD iﬀ −abc
is not a square.
d) If n ≥ 5, Rf is a UFD.
Remark: If f (x1 , . . . , x4 ) s a diagonal quadratic form, we may permute and/or
rescale the coeﬃcients so that either condition (i) or condition (ii) of part c) holds.
Proof. a) If n ≤ 2, Rf is never an integrally closed domain.
b) The proof of Theorem 376 goes through to show that if f is isotropic (i.e., not
anisotropic), Rf is not a UFD. The anisotropic case is due to Samuel [Sa64].
Part c) is due to T. Ogoma [O74].
Part d) goes back at least to van der Waerden [vdW39]. In [Nag57], M. Nagata
gives a short proof using Theorem 375.
It is also interesting to consider aﬃne rings of inhomogeneous quadric hypersurfaces.
For instance, we state without proof the following result.
Theorem 378. For n ≥ 1, let Rn := R[t1 , . . . , tn+1 ]/(t2 + . . . + t2 +1 − 1) be the
n
1
ring of polynomial functions on the nsphere S n .
a) (Bouvier [Bou78]) If n ≥ 2, then Rn is a UFD.
b) (Trotter [T88]) R1 is isomorphic to the ring R[cos θ, sin θ] of real trigonometric
polynomials, in which (sin θ)(sin θ) = (1+cos θ)(1 − cos θ)) is an explicit nonunique
factorization into irreducible elements. Hence R1 is not a UFD.
15.11. AuslanderBuchsbaum.
In the realm of algebraic geometry, the following is perhaps the single most important result about UFDs:
Theorem 379. (AuslanderBuchsbaum) A regular local ring is a UFD. COMMUTATIVE ALGEBRA 219 A Noetherian local ring R with maximal ideal m is regular if the dimension of
m/m2 as an R/mvector space is equal to the Krull dimension of R. This is an
algebraic expression of the geometric property of having a welldeﬁned tangent
space at the point m. If, for instance, R = C[x1 , . . . , xn ]/I is the coordinate ring of
an aﬃne complex variety, then maximal ideals correspond to points Pm of the zero
set V (I ) ⊂ Cn , and regularity of the localization at m occurs iﬀ the corresponding
topological space has the structure of a smooth manifold in some neighborhood
around the point Pm .
´
16. Principal rings and Bezout domains
A ring R in which every ideal is principal is called a principal ring. If R is
moreover a domain, it is called a principal ideal domain (PID).
16.1. Principal ideal domains.
We begin with the following result, essentially a restatement of X.X in §X.X .
Proposition 380. a) A PID has dimension at most one.
b) A UFD is a PID iﬀ it has dimension at most one.
Proof. a) Any domain has positive Krull dimension. Moreover, if (x) ⊂ (y ) are
two nonzero principal prime ideals, then we may write x = cy for some 0 ̸= c ∈ R.
Since x is prime, either x  c or x  y . If dx = c, then dcy = dx = c, and cancellation
gives that y is a unit, contradiction. Therefore y = dx = dcy and cancellation gives
dc = 1, i.e., c, d ∈ R× and (x) = (y ).
b) We recall only that by Kaplansky’s Theorem 343, any nonzero ideal in a UFD
contains a prime element.
Exercise X.X: Let R be a PID and let S ⊂ R \ {0} be a multiplicative subset. Show
that the localization S −1 R is a PID (which is a ﬁeld iﬀ S ∩ R× ̸= ∅).
Theorem 381. A domain in which each prime ideal is principal is a PID.
Proof. Suppose not. Then the set of all nonprincipal ideals is nonempty. Let {Ii }
∪
be a chain of nonprincipal ideals and put I = i Ii . If I = (x) then x ∈ Ii for some
i, so I = (x) ⊂ Ii implies I = Ii is principal, contradiction. Therefore by Zorn’s
Lemma there exists an ideal I which is maximal with respect to the property of
not being principal. As is so often the case for ideals maximal with respect to some
property or other, we can show that I is necessarily prime. Indeed, suppose that
xy ∈ I but neither x nor y lies in I . Then the ideal J = ⟨I, x⟩ is strictly larger
than I , hence principal: say J = (d). Let I ′′ = (I : x) be the set of all elements
of r of R such that rx ∈ I . Then I ′′ is an ideal containing I and y , hence strictly
larger than I and thus principal: say I ′′ = (c). Let r ∈ I , so r = ud. Now u(d) ⊂ I
so ux ∈ I so u ∈ I ′′ . Thus we may write u = vc and r = vcd. This shows I ⊂ (cd).
Conversely, c ∈ I ′′ implies cx ∈ I so c(I + xR) = cJ ⊂ I so cd ∈ I . Therefore
I = (cd) is principal, contradiction.
We now follow with some very familiar examples.
Proposition 382. The integer ring Z is a principal ideal domain. 220 PETE L. CLARK To be honest, I can hardly imagine a reader without prior knowledge of this result
(a staple of undergraduate mathematics) who has made it this far. Therefore, with
our tongue slightly in cheek, we present a “structural” proof. To make clear the
rules of our little game, let us say in advance that the only “prior knowledge” we
will require about the ring Z is that for any positive integer n, the quotient Z/nZ
has (ﬁnite!) order n. To show this, let z be any nonzero integer; note that the
set of integers k such that z − kn ≥ 0 is bounded above so therefore has a largest
element K ; and thus 0 ≤ z − Kn < n.
Lemma 383. The integer ring Z is a unique factorization domain.
Proof. Concretely, we claim that for every integer n > 1, there exist not necessarily
distinct prime numbers p1 , ..., pr such that n = p1 · · · pr and also that if we have
any s prime numbers q1 , . . . , qs such that n = q1 · · · qs , then r = s and there exists
a permutation σ of {1, . . . , r} such that for all 1 ≤ i ≤ r we have qi = pσ(i) .
Here are two proofs which the reader may not have seen before:
First proof: Indeed a decomposition n = p1 · · · pr corresponds to a composition
series for the Zmodule Z/nZ. Since Z/nZ is ﬁnite, it is certainly Noetherian
and Artinian, so composition series exist. Moreover the JordanH¨lder theorem
o
implies that any two composition series have the same number of terms – i.e.,
r = s = ℓ(Z/nZ) – and that after a permutation the sequences of isomorphism
classes of composition factors become identical.
Second proof (Lindemann [Li33], Zermelo [Ze34]): We prove both the existence
and uniqueness of the factorization by an inductive argument, speciﬁcally by appeal to the wellordering of the positive integers under ≤.
Existence: let S be the set of integers n > 1 which do not have at least one
prime factorization. We wish to show that S is empty so, seeking a contradiction,
suppose not. Then by wellordering S has a least element, say N . If N is prime,
then we have found a prime factorization, so suppose it is not prime: that is, we
may write N = N1 N2 with 1 < N1 , N2 < N . Thus N1 and N2 are too small to lie
in S so each have prime factorizations, say N1 = p1 · · · pr , N2 = q1 · · · qs , and then
N = p1 · · · pr q1 · · · qs gives a prime factorization of N , contradiction!
Uniqueness: we claim that the factorization of a positive integer is unique. Assume not; then the set of positive integers which have at least two diﬀerent standard
form factorizations is nonempty, so has a least element, say N , where:
(20) N = p1 · · · pr = q1 · · · qs . Here the pi ’s and qj ’s are prime numbers, not necessarily distinct from each other.
However, we must have p1 ̸= qj for any j . Indeed, if we had such an equality, then
we could cancel and, by an inductive argument we have already rehearsed, reduce
to a situation in which the factorization must be unique. In particular p1 ̸= q1 .
Without loss of generality, assume p1 < q1 . Then, if we subtract p1 q2 · · · qs from
both sides of (20), we get
(21) M := N − p1 q2 · · · qs = p1 (p2 · · · pr − q2 · · · qs ) = (q1 − p1 )(q2 · · · qs ). By the assumed minimality of N , the prime factorization of M must be unique.
However, (21) gives two diﬀerent factorizations of M , and we can use these to get a COMMUTATIVE ALGEBRA 221 contradiction. Speciﬁcally, M = p1 (p2 · · · pr − q2 · · · qs ) shows that p1  M . Therefore, when we factor M = (q1 − p1 )(q2 · · · qs ) into primes, at least one of the prime
factors must be p1 . But q2 , . . . , qj are already primes which are diﬀerent from p1 ,
so the only way we could get a p1 factor is if p1  (q1 − p1 ). But this implies p1  q1 ,
and since q1 is also prime this implies p1 = q1 . Contradiction!
Now we come to the proof of Proposition 382. Applying the lemma together with
Kaplansky’s Theorem (Theorem 343), we see that every nonzero prime ideal p of Z
contains a prime element, say p. But the quotient Z/(p) is a ﬁnite integral domain,
hence a ﬁeld, hence (p) is maximal and therefore (p) = p. This shows that every
prime ideal of Z is principal, and now ﬁnally we apply Theorem 381.
Proposition 384. For any ﬁeld k , the ring R = k [t] is a principal ideal domain.
Proof. By Theorem 361, R is a UFD. The remainder of the argument is quite
similar to the case of R = Z. Namely it suﬃces to show that the ideal generated
by any prime element is maximal. But a prime element of k [t] is, among other
things, a polynomial p(t) of positive degree d, and the quotient R/(p(t)), being
ﬁnite dimensional over a ﬁeld k , is an Artinian integral domain. Thus R/(p(t)) is
a ﬁeld and (p(t)) is maximal.
Conversely:
Exercise X.X: Let R be a ring, and suppose that the univariate polynomial ring
R[t] is a PID. Show that R is a ﬁeld.
Proposition 385. Let R be a Noetherian local ring with a principal maximal ideal
m = (a). Then every nonzero ideal of R is of the form (ai ) for some i ∈ N. In
particular, R is a principal ring.
∩
∩
Proof. The key is the Krull intersection theorem: i mi = i (ai ) = 0. It follows
that for any nonzero r ∈ R, there exists a largest i ∈ N such that r ∈ (ai ), i.e., there
exists s ∈ R such that r = sai . But if s were not a unit then it would lie in m and
thus r would lie in mi+1 , contradiction. So s is a unit and (r) = mi . Thus to every
nonzero element r of I we attach a nonnegative integer ir . Now if I is any nonzero
ideal of R, choose a nonzero element r of I with ir minimal among elements of I .
Then I ⊃ (r) = mi , and the other containment follows by minimality of ir .
Corollary 386. For any ﬁeld k , the formal power series ring k [[t]] is a PID.
Proof. We have seen that a power series ring over a domain is a domain. By
Theorem 218b), k [[t]] is Noetherian. Quite generally, for a ring R the units of R[[t]]
are precisely those formal power series f = a0 + a1 t + . . . with a0 ∈ R× ; since
k is a ﬁeld this means that the units are those with nonzero constant term. The
complement of the set of units is the principal ideal (t), so k [[t]] satisﬁes all the
hypotheses of Proposition 385 and is therefore a PID.
Remark: If we had available the theory of completions, we could deduce Corollary
386 directly from Proposition 384.
16.2. A word on Euclidean functions. 222 PETE L. CLARK We cannot quite allow ourselves to leave the topic of the previous section without making some mention of the standard way to show that rings like Z and k [t]
are PIDS. Namely, if R is an integral domain, then a Euclidean function is a
function N : R \ {0} → N such that: for all a ∈ R and b ∈ R \ {0}, there exists
q, r ∈ R such that a = qb + r with r = 0 or N (r) < N (b).
Exercise: For each of the following rings R, we give a function N : R \ {0} → N.
Verify that N is a Euclidean function on R.
a) R = Z, N (a) = a.
b) R = k [t], N (p∑ = deg(p(t)).
(t))
∞
c) R = k [[t]], N ( n=0 an tn ) := the least n such that an ̸= 0. √
√
d) For d ∈ {−2, −1, 2, 3}, R = Z[ d] = Z[t]/(t2 − d), N (a + b d) = a2 − db2 .
Exercise: Let R be a local PID. Use Proposition 385 to construct a Euclidean
function on R.
The virtue of Euclidean functions lies in the following result:
Proposition 387. A domain R which admits a Euclidean function is a PID.
Proof. Let I be any nonzero ideal of R, and choose an element x in I such that
N (x) is minimal. For any y ∈ I , by deﬁnition of a Euclidean function we may
write y = qx + r with r = 0 or N (y ) < N (x). Our choice of x rules out the second
alternative and thus we have y = qx, i.e., y ∈ (x).
Remark: It is common to deﬁne a Euclidean domain as a domain which admits
some Euclidean function. In my opinion, this deﬁnition is somewhat of a false step
in the theory. The merit of the notion of a Euclidean function is that for certain
rings R there is an evident Euclidean function (e.g. for Z and k [t]), and when
this occurs one deduces immediately that R is a PID. Moreover, if the Euclidean
function is (in some sense) computable, the Euclidean algorithm can be used to
ﬁnd greatest common divisors and, accordingly, generators of ideals.
In contrast, since the speciﬁc Euclidean function is not part of the deﬁnition of a
Euclidean domain, in order to show that a given ring R is not Euclidean one needs
to argue that no Euclidean function can exist. Evidently if R is not a PID, no such
function can exist. However, there are PIDs which admit no Euclidean function
(more details are given below).
One way to assess the naturality of the notion Euclidean domain is to examine
its stability under small perturbations of the deﬁnition. In other words, to what
extent do similar axioms for a “Euclidean function” lead to an equivalent class of
rings?
There are several results to the eﬀect that a certain weakening of the deﬁnition
of a Euclidean function captures precisely the class of all PIDs. For example:
Theorem 388. (DedekindHasse) For a domain R with fraction ﬁeld K , TFAE:
(i) There exists a function N : K → Q satisfying:
(QN1) ∀x ∈ K, N (x) ≥ 0; N (x) = 0 ⇐⇒ x = 0.
(QN2) ∀x, y ∈ K, N (xy ) = N (x)N (y ).
(QN3) ∀x ∈ R, N (x) ∈ Z. COMMUTATIVE ALGEBRA 223 (QN4) ∀x ∈ R, N (x) = 1 ⇐⇒ x ∈ R× .
(QN5) ∀x ∈ K \ R, ∃ a, b ∈ R such that 0 < N (ax − b) < 1.
(ii) R is a principal ideal domain.
Exercise: Prove Theorem 388. (Suggestions: that (i) =⇒ (ii) is the usual argument that an ideal is generated by any element of minimal norm. Conversely, if R is
a PID, deﬁne N on R \{0} by, e.g., setting N (x) to be 2r , where r = ℓ(R/xR) is the
number of irreducibles appearing in a factorization of x and extend this map to K .)
In another direction, P. Samuel considered the notion of a WEuclidean function
on a domain R. Here W is a wellordered set and N : R → W is a function such
that for all a ∈ R, b ∈ R \ {0} such that b a, ∃ q, r ∈ R with a = qb + r and
N (r) < N (b). If R admits, for some W , a W Euclidean function, then R is a PID.
Exercise: Show that a domain is W Euclidean for some ﬁnite W iﬀ it is a ﬁeld.
Say that a domain is SamuelEuclidean if it is W Euclidean for some √
wellordered
set W . Samuel remarks that the an imaginary quadratic ﬁelds Q( −d) has a
SamuelEuclidean ring of integers Rd iﬀ d = 1, 2, 3, 7, 11. On the other hand, it
goes back at least to Gauss that for each of d = 19, 43, 67, 163 the ring Rd is a PID.
Thus there are PIDs which are not SamuelEuclidean. Samuel further showed that
any SamuelEuclidean ring is W Euclidean for a unique minimal wellordered set
(up to canonical order isomorphism) WR and asked the question of whether one has
WR ≤ N for all domains R. This was answered in the negative by Hiblot [Hi75].
16.3. Some structure theory of principal rings.
Proposition 389. Let R be a principal ring.
a) For any multiplicative subset S , the localization RS is principal.
b) For any ideal I of R, the quotient R/I is principal.
Exercise X.X: Prove Proposition 389.
Proposition 390. Let R1 , . . . , Rn be ﬁnitely many rings. TFAE:
(i) Each Ri is a principal ring.
∏n
(ii) The direct product i=1 Ri is a principal ring.
Exercise X.X: a) Prove Proposition 390.
b) Exhibit an inﬁnite direct product of PIDs which is not a principal ring.
Deﬁnition: A principal ring (R, m) is special if it is a local Artinian ring, i.e.,
if it is local and the maximal ideal is principal and nilpotent. The complete structure of ideals in special principal rings can be deduced from Proposition 385: if n
is the least positive integer such that mn = 0, then the ideals of R are precisely the
powers mi = (π i ) for 0 ≤ i ≤ n.
We can now state a structure theorem for principal rings, which appears in Zariski
and Samuel (so far as I know, for the ﬁrst time).
Theorem 391. (ZariskiSamuel) Every principal ring is the direct product of a
ﬁnite number of PIDs and special principal rings.
Much of the work of the proof is contained in the following preparatory result: 224 PETE L. CLARK Lemma 392. Let R be a principal ring.
a) Let p q be prime ideals of R. Then q contains no prime ideals other than itself
and p and every primary ideal contained in q contains p.
b) If p is a nonmaximal prime ideal and a ⊂ p is a primary ideal, then a = p.
c) Any two incomparable prime ideals of R are comaximal.
Proof. a) Suppose that we have p
q and also q
r, where r is a prime ideal.
Write p = (p), q = (q ) and r = (r). There exists a ∈ R such that p = aq . Since
p is a prime element and p does not divide q , we must have p  a, so that there
exists b ∈ R with a = pb and thus p = pbq . Now since p(1 − bq ) = 0 ∈ r ⊂ q and
1 − bq ∈ R \ q, r does not contain 1 − bq and therefore we must have p ∈ r, i.e.,
p ⊂ r ⊂ q. But now modding out by p we get 0 ⊂ r/p ⊂ q/p in the principal ideal
domain R/p, which is a onedimensional ring, and therefore r = p or r = q.
b) . . .
c) Now let p1 and p2 be incomparable prime ideals. If p1 is maximal, then it is
comaximal with any prime ideal which is not contained in it. Similarly, p2 is not
maximal either. So if they are not comaximal, there exists a maximal ideal q strictly
containing both p1 and p2 . But applying part a) to p1 q gives a contradiction.
Proof of Theorem 391: Let
(0) = n
∩ ai i=1 be an irredundant primary decomposition of (0), and let pi = rad(ai ). Then the pi ’s
are pairwise comaximal; otherwise, for any distinct {i, j }, by Lemma 392 we would
have, say, pi pj and then part b) of the lemma gives pi = ai ⊂ aj , contradicting
irredundancy. Since the radicals of the ai ’s are pairwise comaximal, so are the ai ’s
themselves (Proposition XX). So we may apply the Chinese Remainder Theorem,
getting
n
n
∩
∏
R = R/(0) = R/
ai =
R/ai .
i=1 i=1 Each factor is, of course, a principal ring. Fix any 1 ≤ i ≤ n.
Case 1: Suppose ﬁrst that pi is maximal. If ai were contained in any other prime
ideal p, then rad(ai ) = pi ⊂ rad(p) = p, contradiction. So R/ai is a special PIR.
Case 2: Otherwise, pi = ai , so that R/ai is a PID.
We quote without proof the following somewhat stronger result:
Theorem 393. (Hungerford structure theorem [Hun68])
For a commutative ring R, TFAE:
(i) R is a principal ring.
∏n
(ii) R ∼ i=1 Ri , and each Ri is a quotient of a PID.
=
Exercise: Derive Theorem 391 from Theorem 393.
Exercise: Show that for a commutative ring R, TFAE:
(i) R is a special principal ring.
(ii) R is the quotient of a DVR by a nonzero ideal.
Exercise: Let k be any ﬁeld. Let R = k [x, y ], and m be the maximal ideal ⟨x, y ⟩ COMMUTATIVE ALGEBRA 225 in R. Show that the quotient ring S = R/m2 is nonprincipal. In particular, if k is
ﬁnite, then S is a ﬁnite nonprincipal local ring.
16.4. B´zout domains.
e
Proposition 394. Let a, b be elements of a domain. If the ideal ⟨a, b⟩ is principal,
then its generator is a greatest common divisor of a and b.
Proof. In other words, we are assuming the existence of some d ∈ R such that
dR = aR + bR. Then a, b ∈ dR, so d is a common divisor of a and b. If e  a and
e  b then e ∈ aR + bR = dR, so e  d.
Corollary 395. For a domain R, TFAE:
(i) Every ﬁnitely generated ideal is principal.
(ii) For any two elements a and b of R, gcd(a, b) exists and is an Rlinear combination of a and b.
Proof. (i) =⇒ (ii) is immediate from Proposition XX: gcd(a, b) will be a generator
of the ideal ⟨a, b⟩. Conversely, if d = gcd(a, b) exists and is of the form d = xa + yb
for some x, y ∈ R, then clearly (d) = ⟨a, b⟩, so that every ideal with two generators
is principal. By an obvious induction argument, we conclude that any ﬁnitely
generated ideal is principal.
´
At least according to some, it was Etienne B´zout who ﬁrst explicitly noted that
e
for polynomials P, Q ∈ k [t], gcd(a, b) exists and is a linear combination of a and
b: this fact is called B´zout’s identity or B´zout’s Lemma. For this (somee
e
what tenuous) reason, a possibly nonNoetherian domain satisfying the equivalent
conditions of Corollary 395 is called a B´zout domain.
e
Theorem 396. For a B´zout domain R, the following are equivalent:
e
(i) R is a PID.
(ii) R is Noetherian.
(iii) R is a UFD.
(iv) R is an ACCP domain.
(v) R is an atomic domain.
Proof. (i) ⇐⇒ (ii) immediately from the deﬁnitions.
(i) =⇒ (iii): this is Corollary 344.
(iii) =⇒ (iv) =⇒ (v) holds for all domains.
(v) =⇒ (iii): A B´zout domain is a GCDdomain is an ELdomain, so a B´zout
e
e
atomic domain is a UFD.
(iv) =⇒ (ii): assume that R is not Noetherian. Then it admits an ideal I which
is not ﬁnitely generated, which we can use to build an inﬁnite strictly ascending
chain of ﬁnitely generated ideals I1 I2 . . . I . But since R is B´zout, each Ii
e
is principal, contradicting ACCP.
Say that a domain is properly B´zout if it is B´zout but not a PID.
e
e
Recall that we have already seen some interesting examples of properly B´zout
e
domains, namely the ring of entire functions (Theorem 147) and the ring of all
algebraic integers (Theorem 129).
To see many more examples of B´zout domains (including properly B´zout doe
e
mains), we move on to the next topic: valuation rings. 226 PETE L. CLARK 17. Valuation rings
17.1. Basic theory. Consider the divisibility relation – i.e., a  b – on a domain
R. Evidently it is reﬂexive and transitive, so is a quasiordering.57 Divisibility
need not be a partial ordering because a  b and b  a does not imply that a = b
but only that a and b are associates: (a) = (b). However, one of the ﬁrst ideas
of ideal theory is to view associate elements as being somehow “equivalent.” This
motivates us to consider the equivalence relation on R in which a ∼ b iﬀ (a) = (b).
This is easily seen to be a monoidal equivalence relation. In plainer language,
if (a1 ) = (a2 ) and (b1 ) = (b2 ), then (a1 b1 ) = (a2 b2 ). We can therefore consider the
commutative monoid of principal ideals of R under multiplication, on which the
divisibility relation is a partial ordering.
Having made a quasiordering into a partial ordering, it may be natural to ask for
conditions under which the divisibility relation induces a total ordering. Equivalently, for any a, b ∈ R either a  b or b  a.
Proposition 397. Let R be a domain with fraction ﬁeld K . TFAE:
(i) For every a, b ∈ R, a  b or b  a.
(ii) For every 0 ̸= x ∈ K , x ∈ R or x−1 ∈ R.
Exercise: Prove Proposition X.X.
A domain R satisfying the conditions of Proposition 397 is called a valuation
domain or valuation ring.
Note that any ﬁeld is a valuation ring. This is a trivial example which is often
implicitly excluded from consideration (we will try our best to be explicit in our
exclusion of trivial cases). Apart from this, in a ﬁrst algebra course one may not
see examples of valuation rings. But we have: if p is a prime number, then the ring
y
Z(p) of integers localized at p is such an example. Deﬁne x p y if ordp ( x ) ≥ 0. Then
pdivisibility is immediately seen to be a total quasiordering: given two integers,
at least one pdivides the other. The fundamental theorem of arithmetic implies
x  y ⇐⇒ ∀ primes p, x p y.
However, in Z(p) , we have x  y ⇐⇒ x p y , i.e., we have localized the divisibility
relation to get a total quasiorder: Z(p) is a valuation domain.
This argument generalizes as follows: let R be a PID58 and p = (π ) be a prime
ideal of R. We deﬁne ordp (x) to be the least n such that (x) ⊃ pn , and extend it
to a map on K × by ordp ( x ) = ordp (x) − ordp (y ). (One should check that this is
y
y
welldeﬁned; this is easy.) Finally, we deﬁne x p y to mean ordp ( x ) ≥ 0. Arguing
as above, we see that the localization Rp is a valuation ring.
Note that in showing that Rp was a valuation domain we proceeded by constructing
a map ordp on the nonzero elements of the fraction ﬁeld K . This can be generalized,
as follows: if R is a domain with quotient ﬁeld K , we can extend the divisibility
y
y
relation to K × by saying that x  y iﬀ x ∈ R. Clearly x  y and y  x iﬀ x is a unit
57By deﬁnition, a quasiordering is a reﬂexive, transitive binary relation on a set.
58In fact we can take R to be any Dedekind domain, as soon as we know what such a thing is. See §18. COMMUTATIVE ALGEBRA 227 in R. Therefore the quotient of (K × , ·) on which divisibility (from R!) becomes a
partial ordering is precisely the quotient group K × /R× .
y
For [x], [y ] ∈ K × /R× , let us write [x] ≤ [y ] if [ x ] ∈ R. (Take a second and
check that this is welldeﬁned.) Exercise X.X: Show that the divisibility quasiordering on R is a total ordering
iﬀ the ordering on K × /R× is a total ordering.
In other words, if R is a valuation ring, then the canonical map v : K × → K × /R×
is a homomorphism onto a totally ordered abelian group. Let us relabel the quotient group by G and denote the group law by addition, so that the homomorphism
property gets recorded as
(VRK1) ∀x, y ∈ K × v (ab) = v (a) + v (b).
We recover R as R = {x ∈ K ×  v (x) ≥ 0} ∪ {0}.
Everything that has been said so far takes into account only the multiplicative
structure on R. So the following additional property is very important:
(VRK2) ∀x, y ∈ K ×  x + y ̸= 0, v (x + y ) ≥ min(v (x), v (y )).
Indeed, suppose WLOG that v (x) ≤ v (y ), i.e.,
v (x) ≤ v (x + y ). y
x ∈ R. Then x+y
x = 1+ y
x ∈ R so Exercise: Suppose that v (x) ̸= v (y ). Show that v (x + y ) = min(v (x), v (y )).
Let (G, +, ≤) be a totally ordered abelian group. We write G+ = {g ∈ G  g ≥ 0},
so G+ is a totally ordered submonoid of G. A (Gvalued) valuation on a ﬁeld
K is a surjective map v : K × → G satisfying (VRK1) and (VRK2) above.
Exercise: Let v : K × → G be a valuation. Let R be the set of elements of K ×
with nonnegative valuation, together with 0. Show that R is a valuation ring with
fraction ﬁeld K .
Exercise: Let R be a domain, G a totally ordered group and v : R \ {0} → G+ be
a map which satisﬁes both of the following properties:
(VRR1) ∀x, y ∈ R \ {0}, v (ab) = v (a) + v (b).
(VRR2) ∀x, y ∈ R \ {0}  x + y ̸= 0, v (x + y ) ≥ min(v (x), v (y )).
(VRR3) v (R \ {0}) ⊃ G+ .
Show that there is a unique extension of v to a valuation v : K × → G, namely
v (x/y ) = v (x) − v (y ).
Proposition 398. A valuation ring is a local domain, in which the unique maximal
ideal m consists of elements of positive valuation.
Proof. The set of elements of strictly positive valuation form an ideal m of R. Its
complement, the set of elements of zero valuation, is the group of units. 228 PETE L. CLARK Proposition 399. Let I be a ﬁnitely generated ideal in a valuation ring. Then the
set v (I ) has a least element, and for any x ∈ I of minimal valuation, I = v (x).
Proof. Write I = ⟨x1 , . . . , xn ⟩ with v (x1 ) ≤ . . . ≤ v (xn ). Then for any r1 , . . . , rn ∈
R, v (r1 x1 + . . . + rn xn ) ≥ mini v (ri xi ) ≥ v (x1 ), so x1 is an element of I of minimal
xi
xi
valuation. Thus for all i > 1, v ( x1 ) ≥ 0, so x1 ∈ R and x1  xi . So I = ⟨x1 ⟩.
Corollary 400. A valuation ring is a B´zout domain. In particular, a Noetherian
e
valuation ring is a PID.
Conversely:
Proposition 401. A local B´zout domain is a valuation domain.
e
Proof. Let x, y be elements of a local B´zout domain, and suppose d = gcd(x, y ).
e
Then x and y are coprime. Since in a local ring the nonunits form an ideal, this
d
d
implies that at least one of x and y is a unit. In other words, up to associates,
d
d
d = x (so x  y ) or d = y (so y  x).
A valued ﬁeld is a ﬁeld together K together with a valuation v : K × → G. An
isomorphism of valued ﬁelds (K, v ) → (K ′ , v ′ ) is an isomorphism f : K → K ′ of
ﬁelds such that v = v ′ ◦ f . A valued ﬁeld is trivial if G is the trivial group. Evidently each ﬁeld carries a unique trivial valuation up to isomorphism: the valuation
ring is K itself. The reader will lose nothing by making the tacit assumption that
all valuations are nontrivial.
Exercise: A chain ring is a ring R in which the partially ordered set I (R) of
ideals of R is linearly ordered.
a) Show that for a ring R, TFAE:
(i) R is a chain ring.
(ii) For all x, y ∈ R, either xR ⊂ yR or yR ⊂ xR.
b) Show that a domain R is a chain ring iﬀ it is a valuation ring.
17.2. Ordered abelian groups.
Let (G, +) be an abelian group, written additively. In particular the identity element of G will be denoted by 0. As for rings, we write G• for G \ {0}.
By an ordering on G we mean a total (a.k.a. linear) ordering ≤ on G which
is compatible with the addition law in the following sense:
(OAG) For all x1 , x2 , y1 , y2 ∈ G, x1 ≤ x2 and y1 ≤ y2 implies x1 + y1 ≤ x2 + y2 .
One has the evident notions of a homomorphism of ordered abelian groups, namely
an isotone group homomorphism.
Exercise: Let (G, ≤) be an ordered abelian group.
a) Let x ∈ G• . Show that either x > 0 or −x > 0 but not both.
b) Show that for all x, y ∈ G, x ≤ y ⇐⇒ −y ≤ −x.
Exercise: Let (G, ≤) be an ordered abelian group, and let H be a subgroup of
G. Show that the induced order on H makes H into an ordered abelian group. COMMUTATIVE ALGEBRA 229 Example: For any ordered ﬁeld (F, ≤), the additive group (F, +) is an ordered
abelian group. In particular, the additive group (R, +) of the real numbers is an
ordered abelian group, as is any subgroup. In particular, (Z, +) and (Q, +) are
ordered abelian groups.
Exercise: Exhibit an abelian group which admits two nonisomorphic orderings.
Example (Lexicographic ordering): Let {Gi }i∈I be a nonempty indexed family
of ordered abelian groups. Suppose that we are given a wellordering on the index
∏
set I . We may then endow the direct product G = i∈I Gi with the structure of
an ordered abelian group, as follows: for (gi ), (hi ) ∈ G, we decree (gi ) < (hi ) if for
the least index i such that gi ̸= hi , gi < hi .
Exercise: Check that the lexicographic ordering on the product
a total ordering on G. ∏
i ∈I Gi is indeed Theorem 402. (Levi [Lev43]) For an abelian group G, TFAE:
(i) G admits at least one ordering.
(ii) G is torsionfree.
Proof. (i) =⇒ (ii) Suppose < is an ordering on G and let x ∈ G• . Then exactly
one of x, −x is positive; without loss of generality say it is x. Then for all n ∈ Z+ ,
nx = x + . . . x (n times) must be positive, so x has inﬁnite order in G.
(ii) =⇒ (i): Let G be a torsionfree abelian group. By Corollary 93, G is a ﬂat
Zmodule. Tensoring the injection Z → Q gives us an injection G → G ⊗ Q. Since
⊕
Q is a ﬁeld, the Qmodule G ⊗ Q is free, i.e., it is isomorphic to i∈I Q. Choose a
total ordering on I . Give each copy of Q its standard ordering as a subﬁeld of R and
⊕
put the lexicographic ordering on i∈Q Q ∼ G ⊗ Q. Via the injection G → G ⊗ Q
=
this induces an ordering on G.
Exercise X.X: a) Show that the abelian group Z admits exactly one ordering (here
when we say “ordering”, we always mean “ordering compatible with the group
structure in the sense of (OAG).
b) Give an example of an abelian group which admits two distinct – even nonisomorphic – orderings.
An ordered abelian group (G, +) is Archimedean if for all x, y ∈ G with x > 0,
there exists n ∈ Z+ with nx > y .
Exercise X.X:
a) Suppose H is a subgroup of the Archimedean ordered group (G, +). Show that
the induced ordering on G is Archimedean.
b) Let (G, +) be an ordered abelian group such that there exists an embedding
(G, +) → (R, +) into the additive group of the real numbers. Deduce that G is
Archimedean.
Conversely:
o
o
Theorem 403. (H¨lder [H¨01]) Let (G, +) be an ordered abelian group. If G is
Archimedean, there exists an embedding (G, +) → (R, +). 230 PETE L. CLARK Proof. We may assume G is nontrivial. Fix any positive element x of G. We will
construct an order embedding of G into R mapping x to 1.
Namely, let y ∈ G. Then the set of integers n such that nx ≤ y has a maximal
element n0 . Put y1 = y − n0 x. Now let n1 be the largest integer n such that
nx ≤ 10y1 : observe that 0 ≤ n1 < 10. Continuing in this way we get ∑ of integers
a set
∞
n
n1 , n2 , . . . ∈ {0, . . . , 9}. We deﬁne φ(y ) to be the real number n0 + k=1 10k . It is
k
′
′
not hard to show that φ is isotone – y ≤ y =⇒ φ(y ) ≤ φ(y ) – and also that φ is
injective: we leave these tasks to the reader.
But let us check that φ is a homomorphism of groups. For y ∈ G, and r ∈ Z+ ,
n
let 10r be the rational number obtained by truncating φ(y ) at r decimal places.
The numerator n is characterized by nx ≤ 10r y < (n + 1)x. For y ′ ∈ G, if
n′ x ≤ 10r y ′ ≤ (n′ + 1)x, then
(n + n′ )x ≤ 10r (y + y ′ ) < (n + n′ + 2)x,
so
φ(y + y ′ ) − (n + n′ )10−r < 2
10r and thus
4
.
10r
Since r is arbitrary, we conclude φ(y + y ′ ) = φ(y ) + φ(y ′ ).
φ(y + y ′ ) − φ(y ) − φ(y ′ ) < A nontrivial ordered abelian group which can be embedded in R is said to have
rank one. This is indeed part of a general deﬁnition of the rank of an ordered
abelian group (so that, e.g., the lexicographic product of n copies of Z has rank n),
but as we have no need for the general concept and the deﬁnition is not immediately
perspicuous, we omit it here.
The following result gives another characterization of valuation rings of rank one.
Proposition 404. For a valuation ring R, TFAE:
(i) R has rank one, i.e., the value group is Archimedean.
(ii) R has Krull dimension one.
Proof. (i) =⇒ (ii): Since the value group is nontrivial, R is not a ﬁeld. Let p
be a prime ideal which is strictly contained in the maximal ideal m of R. Choose
α ∈ m \ p and put x = v (α). Let β ∈ p• and put y = v (β ). Using the Archimedean
n
property of the value group, there exists n ∈ Z+ such that nx > y and thus α ∈ R.
β
It follows that αn ∈ βR ⊂ p, and since p is prime, α ∈ p, contradiction.
(ii) =⇒ (i): Let β ∈ m• . Since m is the unique prime containing β , r(βR) = m:
for all α ∈ m there is n ∈ Z+ with β  αn . So the value group is Archimedean.
Lemma 405. Let R be a domain, (G, ≤) a totally ordered commutative group, and
let G+ = {g ∈ G  g ≥ 0}. Then:
a) G+ is an ordered commutative monoid.
b) The monoid ring R[G+ ] is an integral domain.
c) The group ring R[G] is naturally isomorphic to the localization of R[G+ ] at the
multiplicative subset G+ . In particular, R[G] is an integral domain.
Exercise X.X: a) Write out the statements of Lemma 405 when G = Z.
b) Prove Lemma 405. COMMUTATIVE ALGEBRA 231 Theorem 406. (Malcev, Neumann) For any ordered abelian group G, there exists
a valuation domain with value group isomorphic to G.
Proof. It suﬃces to construct a ﬁeld K and a surjective map v : K × → G satisfying
(VD1K) and (VD2K). Let k be any ﬁeld and put A = k [G≥0 ]. By Lemma 405, A
is a domain; let ∑ be its fraction ﬁeld. Deﬁne a map v : A• → G by sending a
K
nonzero element g∈G ag g to the least g for which ag ̸= 0. It is easy to check that
v satisﬁes (VD1)(VD3) of Exercise X.X above, and therefore extends uniquely to
a valuation v : K × → G, where K is the fraction ﬁeld of R.
Recall from §X.X the notion of a “big monoid ring” k [[Γ]], the collection of all
functions f : Γ → k under pointwise addition and convolution product. As we saw
though, in order for the convolution product to be deﬁned “purely algebraically” –
i.e., without recourse to some limiting process – we need to impose the condition
of divisor ﬁniteness on Γ. It follows easily from Proposition 411 that for Γ = G≥0
the monoid of nonnegative elements in a totally ordered abelian group, divisor
ﬁniteness holds iﬀ G = Z, i.e., iﬀ the valuation is discrete.
However, Malcev [Mal48] and Neumann [Neum49] independently found a way
around this by considering a set in between k [G≥0 ] and k [[G≥0 ]]. Namely, deﬁne
kMN [G≥0 ] to be the set of all functions f : G≥0 → k such that the support of f
– i.e., the set of g ∈ G≥0 such that f (g ) ̸= 0 – is wellordered. It turns out that
on such functions the convolution product can be deﬁned and endows kMN [G≥0 ]
with an integral domain. The fraction ﬁeld kMN (G) is simply the collection of
all functions f : G → k with wellordered support. Moreover, mapping each such
nonzero function to the least element of G in its support gives a Gvalued valuation.
The elements of such rings are called MalcevNeumann series.
17.3. Connections with integral closure.
Let (R, mR ) and (T, mT ) be local rings with R ⊂ T . We say that T dominates
R, and write R ≤ T , if mT ∩ R = mR .
Lemma 407. Let R be a subring of a ﬁeld K , and let p ∈ Spec R. Then there
exists a valuation ring T of K such that R ⊂ T and mT ∩ R = p.
Proof. (Matsumura)
Step 0: We may replace R by Rp and thus asume that (R, p) is a local ring. In this
case, what we are trying to show is precisely that there exists a valuation ring of
K dominating R.
Step 1: Let F be the set of all rings R′ with R ⊂ R′ ⊂ K such that pR′
R′ ,
partially ordered by inclusion. We have R ∈ F , so F ̸= ∅. Moreover the union of a
chain in F is again an element of F , so Zorn’s Lemma gives us a maximal element
T of F . Since pT
T , there exists a maximal ideal m of T containing pT . Since
T ⊂ Tm and Tm ∈ F , by maximality of T we have T = Tm , so (T, m) is a local ring
dominating (Rp , p).
Step 2: We claim that T is a valuation ring.
proof of claim: Let x ∈ K × . We wish to show that at least one of x, x−1 lies in
T . Seeking a contradiction, assume neither is the case. Then T [x] properly contains
T , so by maximality of T we have 1 ∈ pT [x], i.e., we get a relation of the form
1 = a0 + a1 x + . . . + an xn , ai ∈ pT. 232 PETE L. CLARK Since T is local, 1 − a0 ∈ T × , and the relation may be rewritten in the form
(22) 1 = b1 x + . . . + bn xn , bi ∈ m. Among all such relations, we may choose one with minimal exponent n. In exactly
the same way, T [x−1 ] properly contains T and thus there exists a relation
(23) 1 = c1 x−1 + . . . + cn x−m , ci ∈ m, and among all such relations we may choose one with minimal m. Without loss
of generality m ≤ n: otherwise interchange x and x−1 . Then multiplying (23) by
bn xn and subtracting from (22) gives another relation of the form (22) but with
exponent smaller than n, contradiction.
A subring R of a ﬁeld K is a maximal subring if R
intermediate between R and K . K and there is no ring Exercise: a) Show that any ﬁeld K admits at least one maximal subring.
b) Show that if K is not ﬁnite of prime order, then all maximal subrings are nonzero.
c) Excluding the case in which K is ﬁnite of prime order, show that any maximal
subring of K is a valuation ring of K (possibly a ﬁeld).
Exercise: Let K be a ﬁeld, and R a valuation ring of K . Show TFAE:
(i) R is a maximal subring of K .
(ii) R has rank at most one.
Theorem 408. Let K be a ﬁeld and R ⊂ K a subring. Then the integral closure
R of R in K is equal to the intersection of all valuation rings of K containing R.
Proof. Let R be the intersection of all valuation rings of K containing R. Since
each such ring is integrally closed in K and the intersection of a family of rings
each integrally closed in K is again integrally closed in K , R is integrally closed in
K , whence R ⊂ R.
Conversely, let x ∈ K \ R. It suﬃces to ﬁnd a valuation ring of K containing R but
not x. Let y = x−1 . The ideal yR[y ] of R[y ] is proper: for if 1 = a1 y +. . .+an y n with
ai ∈ R, then x would be integral over R. Let p be a maximal ideal of R containing
y . By Lemma 407, there exists a valuation ring T of K such that R[y ] ⊂ K and
mT ∩ R[y ] = p. Then y = x−1 ∈ mT , so x ∈ T .
/
17.4. Another proof of Zariski’s Lemma.
The following result is a close relative of Lemma 407.
Lemma 409. Let K be a ﬁeld and Ω an algebraically closed ﬁeld. Let S be the set
of all pairs (A, f ) with A is a subring of K and f : A → Ω, partially ordered by
(A, f ) ≤ (A′ , f ′ ) ⇐⇒ A ⊂ A′ and f ′ A = f.
Then S contains maximal elements, and for any maximal element (B, g ), B is a
valuation ring of K .
Proof. An easy Zorn’s Lemma argument shows that S has maximal elements.
Let (B, g ) be a maximal element. Put p = Ker(g ); since g (B ) is a subring of the ﬁeld
Ω, it is a domain and thus p is a prime ideal of B . By functoriality of localization,
G extends to a homomorphism Bp → Ω. By maximality of (B, g ) we have Bp = B , COMMUTATIVE ALGEBRA 233 so that B is a local ring with maximal ideal p. If there existed an element x ∈ K
which is transcendental over the fraction ﬁeld of B , then B [x] is a polynomial ring
and certainly g extends to B [x]. So K is algebraic over the fraction ﬁeld of B .
Next let x ∈ K × . We claim that either the ideal pB [x] or pB [x−1 ] is proper.
Indeed this is proved exactly as in Lemma 407 above.
Finally, we show that B is a valuation ring of K . Let x ∈ K • . Without loss of
generality, we may assume that pB [x] is a proper ideal of B (otherwise replace x
by x−1 ). Put B ′ = B [x]. By assumption, pB [x] is contained in a maximal ideal
m of B ′ and m ∩ B = p. Hence the embedding of domains B → B ′ induces an
embedding of ﬁelds k := B/p → B ′ /m = k ′ . Moreover k ′ is generated over k by
the image of the algebraic element x, so k ′ /k is a ﬁnite degree ﬁeld extension. So g
induces an embedding k → Ω, and since Ω is algebraically closed, this extends to
an embedding k ′ = B ′ /m → Ω. By maximality of B , this implies x ∈ B .
Remark: It should be possible to consolidate Lemmas 407 and 409 into a single
result. Let me know if you see how to do it.
Proposition 410. Let A ⊂ B be domains with B ﬁnitely generated as an Aalgebra. Let β ∈ B • . There exists α ∈ A• satisfying the following property: any
homomorphism f of A into an algebraically closed ﬁeld Ω with f (α) ̸= 0 extends to
a homomorphism f : B → Ω with f (β ) ̸= 0.
Proof. Step 0: An easy induction argument on the number of generators reduces
us to the monogenic case: B = A[x].
Step 1: Suppose that x is transcendental over A, i.e., B is a univariate polynomial
ring over A. Write
β = an xn + . . . + a1 x + a0 , ai ∈ A
and put α = a0 . If f : A → Ω is such that f (α) ̸= 0, then since Ω is inﬁnite, there
exists ζ ∈ Ω such that f (an )ζ n + . . . + f (a1 )ζ + f (a0 ) ̸= 0. Using the universal
polynomial of polynomial rings, we may uniquely extend f to a homomorphism
from B to Ω by putting f (x) = ζ , and then f (β ) ̸= 0.
Step 2: Suppose that x is algebraic over the fraction ﬁeld of A. Then so is β −1 .
Hence we have equations of the form
an xm + . . . + a1 x + a0 , ai ∈ A
a′ β −m + . . . + a′ β −1 + a′ , a′ ∈ A.
m
1
0i
Put α = an a′ . Suppose f : A → Ω is any homomorphism with f (α) ̸= 0. We may
m
extend f to a homomorphism from A[α−1 ] → Ω by mapping α−1 to f (α)−1 and
then, by Lemma 409, to a homomorphism f : C → Ω for some valuation ring C
containing A[α−1 ]. By construction x is integral over A[α−1 ]. Since C is integrally
closed, x ∈ C . Thus C contains B and in particular β ∈ C . Similarly, β −1 is
integral over A[α−1 ] so β −1 ∈ C . Thus β ∈ C × , so f (β ) ̸= 0. Restricting to B
gives the desired homomorphism.
Proof of Zariski’s Lemma: Let k be a ﬁeld and B a ﬁeld which is ﬁnitely generated
as a k algebra. We want to show that B is a ﬁnite ﬁeld extension of B . Equivalently,
it is enough to show that B/k is algebraic. In Proposition 410 take A = k , β = 1
and Ω to be an algebraic closure of k .
17.5. Discrete valuation rings. 234 PETE L. CLARK 17.5.1. Introducing DVRs.
Proposition 411. For a valuation ring R with value group G, TFAE:
(i) R is a PID.
(ii) R is Noetherian.
(iii) R is an ACCPdomain.
(iv) G≥0 = {x ∈ G  x ≥ 0} is wellordered.
(v) G is isomorphic to (Z, ≤).
Proof. (i) ⇐⇒ (ii) ⇐⇒ (iii) is a special case of Theorem 396.
(iii) ⇐⇒ (iv): a totally ordered set is wellordered iﬀ there are no inﬁnite strictly
descending chains. But an inﬁnite strictly descending chain in the value group of
R gives rise to an inﬁnite strictly ascending chain of principal ideals in R, and
conversely.
(iv) =⇒ (v): First suppose G is Archimedean, so G → R: this endows G with
a topology. If G is discrete, it is generated by its least positive element hence is
orderisomorphic to Z. If G is not discrete, there exists an inﬁnite strictly decreasing
sequence of positive elements of G converging to 0, so G≥0 is not wellordered. Next
suppose that G is not Archimedean, and choose x, y > 0 such that for all n ∈ Z+ ,
nx < y . Then {y − nx}n∈Z+ is an inﬁnite strictly descending sequence in G≥0 ,
contradicting wellordering.
(v) =⇒ (iv): famously, the standard ordering on Z≥0 is a wellordering.
Exercise X.X: Show directly that a local PID is a valuation ring with value group Z.
A valuation ring satisfying the equivalent conditions of Proposition 411 is called
a discrete valuation ring (or, sometimes, a DVR).
17.5.2. Further characterizations of DVRs.
In many ways, discrete valuation rings are – excepting only ﬁelds – the simplest
class of rings. Nevertheless they have an important role to play in algebra and
arithmetic and algebraic geometry. One reason for this is as follows: every DVR is
a onedimensional Noetherian local ring. The converse does not hold.
Example: Let k be a ﬁeld, and let R be the k subalgebra of k [t] generated by
t2 and t3 . This is a onedimensional Noetherian domain; the ideal m generated by
t2 and t3 is a nonprincipal maximal ideal. Indeed, even in the localization Rm the
ideal mRm is not principal: consider what its order at 0 would be with respect to
the valuation ordt on k (t): it would have to be 1, but there is no such element in Rm .
The question then is to ﬁnd necessary and suﬃcient conditions for a onedimensional
Noetherian local domain to be a DVR. As we have seen, being a PID is enough,
but again, this is not very useful as whether a onedimensional domain is a PID is
diﬃcult to check in practice. Remarkably, it turns out if a local, onedimensional
Noetherian domain has any one of a large number of good properties, it will necessarily be a DVR. Here is the theorem.
Theorem 412. (Recognition Theorem for DVRs) Let R be a onedimensional
Noetherian local domain, with maximal ideal m. TFAE:
m
(i) R is regular: the dimension of m2 as an R/mvector space is 1. COMMUTATIVE ALGEBRA 235 (ii) m is principal.
(iii) R is a PID.
(iv) R is a UFD.
(v) R is integrally closed.
(vi) Every nonzero ideal is of the form mn for some n ∈ N.
Proof. (i) ⇐⇒ (ii): Choose t ∈ π \ π 2 . By assumption, t generates m/m2 , so by
Nakayama’s Lemma t generates m. Conversely, if m is monogenic as an Rmodule,
certainly m/m2 is monogenic as an R/mmodule.
Evidently (iii) =⇒ (ii). Proposition 385 gives (ii) =⇒ (iii) and also (ii) =⇒
(vi). Moreover (iii) ⇐⇒ (iv) by Proposition 380 and (iv) =⇒ (v) by 354. Next,
for all n ∈ N we have (π )n /(π )n+1 ∼ R/m, thus R is regular.
=
(vi) =⇒ (i): Assume that dimR/m m/m2 > 1. Choose u ∈ m \ m2 . Then we
have
m ⟨u, m2 ⟩ m2 .
So we have (i) ⇐⇒ (ii) ⇐⇒ (iii) ⇐⇒ (iv) ⇐⇒ (vi) =⇒ (v).
Finally, we show (v) =⇒ (ii): Let 0 ̸= x ∈ m. Since m is the only prime ideal
containing (x) we must have r((x)) = m. Since R/(x) is Noetherian, its radical,
m/(x), is nilpotent, so there is a unique least n ∈ Z+ such that mn ⊂ (x). Let
y ∈ mn−1 \ (x) and consider the element q = x of the fraction ﬁeld K of R. Since
y
y
y ∈ (x), q −1 = x ∈ R; since R is assumed integrally closed in K , q −1 is not integral
/
/
over R. Then we must have q−1 m is not contained in m, for otherwise m would be
a faithful R[q −1 ]module which is ﬁnitely generated as an Rmodule, contradicting
y
Theorem 310. But by construction, q −1 m = x m ⊂ R, hence q −1 m = R and then
m = Rx = (x).
17.5.3. Modules over DVRs.
Lemma 413. Let R be a DVR with uniformizing element π , and let a ∈ Z+ . Then
the ring Ra = R/(π a ) is selfinjective – i.e., Ra is an injective Ra module.
Exercise X.X: Prove Lemma 413. (Hint: Baer’s Criterion!)
Theorem 414. Let R be a DVR with uniformizing element π , and let M be a
nonzero ﬁnitely generated Rmodule.
a) There is N ∈ N and positive integers n, a1 ≥ a2 ≥ . . . ≥ an such that
(24) M ∼ RN ⊕
= n
⊕ R/(π ai ). i=1 b) The numbers N, n, a1 , . . . , an are invariants of the isomorphism class of the module M : i.e., they are the same for any two decompositions of M as in (24) above.
Proof.
Step 0: Consider the canonical short exact sequence
0 → M [tors] → M → M/M [tors] → 0.
Since M is a ﬁnitely generated module over a Noetherian ring, M [tors] is ﬁnitely
generated. Moreover, M/M [tors] is a ﬁnitely generated torsionfree module over a
PID, hence is free (Proposition 66). Moreover, we know that the rank of a free
module over any (commutative!) ring is welldeﬁned (when R is a domain with
fraction ﬁeld K , the proof is especially easy: the rank of a free module M is 236 PETE L. CLARK dimK M ⊗R K ), so the invariant N in the statement of the theorem is precisely
the rank of M/M [tors]. Moreover, since M/M [tors] is free – hence projective – the
sequence splits, so
M = RN ⊕ M [tors].
We are reduced to the case of a nonzero ﬁnitely generated torsion module M .
Step 1: The annihilator of M is an ideal of R, of which there aren’t so many: it must
be (π a1 ) for some a1 ∈ Z+ . Thus M may be viewed as a faithful Ra1 = R/(π a1 )module. Moreover, choosing an element x of M which is not annihilated by π a1 −1 ,
the unique Ra1 module map Ra1 → M which sends 1 to m is an injection. Taking
M ′ = M/Ra1 , we get a short exact sequence
0 → Ra1 → M → M ′ → 0.
By Lemma 413, Ra1 is an injective Ra1 module, so the sequence splits:
M ∼ Ra ⊕ M ′ .
=
1 Step 2: Since M is ﬁnitely generated over Ra1 , it is a quotient of some Artinian
M
Ra1 module Ra1 , hence by Theorem 188 M is Artinian. Moreover M is a ﬁnitely
generated module over the Noetherian ring, so M is also Noetherian. By Theorem
198, this means that M has ﬁnite length as an Rmodule. Hence so does its direct
summand M ′ and indeed clearly the length of M ′ is less than the length of M .
This completes the proof of part a) by induction.
Step 3: So far we have that a ﬁnitely generated torsion Rmodule is of the form
⊕n
ai
a1
i=1 R/(π ) with positive integers a1 ≥ a2 ≥ . . . ≥ an , and with ann(M ) = (π ).
In order to prove the uniqueness statement of part b), it suﬃces to prove that for
all 0 < b ≤ a, R/(π b ) is an indecomposable R/(π a )module. If so, then
M∼
= n
⊕ R/(π ai ) i=1 is simply the decomposition of the ﬁnite length module M into indecomposables
described in the KrullSchmidt Theorem: in particular, since clearly R/(π a ) ∼
=
R/(π b ) implies a = b (consider annihilators), it is unique up to permutation of the
factors. So suppose that R/(π a ) = M1 ⊕ M2 with M1 , M2 nonzero. If π a does
not annihilate M1 , then as above we can ﬁnd a split embedding R/(π a ) → M1 ,
which contradicts the fact that the length of M1 must be smaller than the length
of R/(π a ). So M1 – and similarly M2 – is annihilated by π a−1 and thus R/(π a )
would be annihilated by π a−1 , a contradiction.
Remark: Theorem 414 is nothing else than the fundamental structure theorem for
modules over a PID in the special case in which the PID has a unique maximal
ideal. But we have not given a proof of this structure theorem for PIDs in these
notes, whereas later we will want to use it to prove a more general structure theorem
for torsion modules over a Dedekind domain. However, in both cases it is easy to
reduce to the local situation, and thus we will get an independent proof.
18. Normalization theorems
We work in the following situation: R is an integrally closed domain with fraction
ﬁeld K , L/K is a ﬁeld extension, and S = IL (R) is the integral closure of R in L.
In more geometric language, S is the normalization of R in the extension L/K .
As above, we may as well assume that L/K is algebraic, since in the general COMMUTATIVE ALGEBRA 237 case, if we let L′ = IK (L) be the algebraic closure of K in L, then S is contained
in L′ anyway. So let us assume this. Then we know that S is integrally closed with
fraction ﬁeld L. We also know that the Krull dimensions of S and R coincide.
The major questions are the following:
(Q1) Is S ﬁnitely generated as an Rmodule?
(Q2) Is S Noetherian?
(Q3) If not, then can anything nice be said about S ?
Note that if R is Noetherian, then an aﬃrmative solution to (Q1) implies an aﬃrmative answer to (Q2). (If R is not Noetherian there is no reason to think that S
should be – e.g. take L = K ! – so we are “only” interested in (Q2) in case R is
Noetherian.) Also, the example of R = Z, K = Q, L = Q above shows that both
(Q1) and (Q2) may have a negative answer if [L : K ] is inﬁnite. So we are most
interested in the ﬁnite case, although, as we shall see, there is something to say in
the inﬁnite case as well.
18.1. The First Normalization Theorem.
The ﬁrst, and easiest, result is the following:
Theorem 415. (First Normalization Theorem) Let R be an integrally closed domain with fraction ﬁeld K , L/K a ﬁnite separable ﬁeld extension, and S = IL (R).
a) There exists a K basis x1 , . . . , xn of L such that S is contained in the Rsubmodule generated by x1 , . . . , xn .
b) Therefore if R is Noetherian, S is ﬁnite as an Rmodule.
c) If R is a PID, then S is a free Rmodule of rank [L : K ].
Proof. By the proof of Proposition 319, for any x ∈ L, there exists 0 ̸= r ∈ R such
that rx ∈ S . Therefore there exists a K basis u1 , . . . , un of L such that ui ∈ S
∑
for all i.59 Now take x ∈ S and write x = i bi ui with bi ∈ K . Since L/K is
separable there are n = [L : K ] distinct K embeddings of L into K , say σ1 , . . . , σn ,
and the discriminant ∆ = ∆(u1 , . . . , un ) = (det(σj (ui )))2 is nonzero. We may put
√
∆) = det(σj (ui )). For all 1 ≤ j ≤ n we have
∑
σj (x) =
bi σj (ui ).
i Using Cramer’s rule, we may solve for the bi to get
∑
∑√
√
∆bi =
dij σj (x), dbi =
ddij σj (x),
j j where the dij ’s are certain polynomials in the σj (ui ) with integer coeﬃcients. This
√
shows that ∆bi and ∆bi are integral over R. Since ∆ ∈ K and R is integrally
closed, we have ∆bi ∈ A. Therefore S is contained in the Rspan ⟨ u1 , . . . , un ⟩R ,
∆
∆
establishing part a). Since over a Noetherian ring, submodules of ﬁnitely genereated
modules are themselves ﬁnitely generated, this implies b). Our work so far shows
that S is a submodule of a free rank n Rmodule, so over an integral domain this
59Note that we have not yet used the separability hypothesis, so this much is true in the case
of an arbitrary ﬁnite extension. 238 PETE L. CLARK implies that S is itself a free Rmodule of rank at most n. Since S ⊗R K = L, the
rank must be exactly n.
This has the following important result, which is the ﬁrst of three fundamental
ﬁniteness theorems in algebraic number theory, the existence of a ﬁnite integral
basis for the ring of integers of any algebraic number ﬁeld:
Corollary 416. Let R = Z, K = Q, L a number ﬁeld of degree n. Then the ring
ZL = Z ∩ K of all algebraic integers lying in L, is an integrally closed, Noetherian
domain of Krull dimension one which is, as a Zmodule, free of rank n.
Proof. The ring ZL is nothing else than IL (Z), so by Proposition 320, it is integrally
closed in its ﬁeld of fractions L. Since Z is a PID and L/Q is ﬁnite separable, Theorem 415 applies to show that ZL ∼ Zn as a Zmodule. Being a ﬁnitely generated
=
Zmodule, still more is it a ﬁnitely generated algebra over the Noetherian ring Z,
so it is itself Noetherian. Since Z, like any PID, has Krull dimension one and ZL is
an integral extension of Z, by Corollary 326 ZL also has Krull dimension one.
A Dedekind domain is an integral domain R which is Noetherian, integrally
closed and of Krull dimension at most one. We will explore this concept in loving
detail later on, but notice for now that Corollary 416 shows in particular that the
ring ZL of integers of any algebraic number ﬁeld L is a Dedekind domain. In fact,
the argument establishes that the normalization S of any Dedekind domain R in a
ﬁnite separable ﬁeld extension L/K is again a Dedekind domain which is ﬁnitely
generated as an Rmodule.
What about the nonseparable case?
Give Kaplansky’s example. COMPLETE ME! ♣
18.2. The Second Normalization Theorem.
Theorem 417. (Second Normalization Theorem) Let R be an integral domain with
fraction ﬁeld K . Suppose that at least one of the following holds:
• R is absolutely ﬁnitely generated i.e., ﬁnitely generated as a Zalgebra – or
• R contains a ﬁeld k and is ﬁnitely generated as a k algebra.
Let L/K a ﬁnite ﬁeld extension. Then S = IL (R) is a ﬁnitely generated Rmodule.
Proof. First spupose that R is a ﬁnitely generated algebra over a ﬁeld k .
Step 0: We may assume that L/K is normal. Indeed, let M be the normal closure
of L/K , so M/K is a ﬁnite normal extension. Let T be the integral closure of R in
M . If we can show that T is ﬁnitely generated over R, then, since R is Noetherian,
the ﬁnitely submodule S is also ﬁnitely generated over R.
Step 1: We will make use of the ﬁeldtheoretic fact that if M/K is normal and L is
the maximal purely inseparable subextension of M/K , then M/L is separable [FT,
§6.3]. Let S be the integral closure of R in L and T the integral closure of R in T .
Then T is moduleﬁnite over R iﬀ T is moduleﬁnite over S and S is moduleﬁnite
over R. Suppose we can show that S is moduleﬁnite over R. Then S is a ﬁnitely
generated Ralgebra so S is a Noetherian integreally closed domain, and the module
ﬁniteness of T over S follows from Theorem 415. Thus we are redueced to the case
in which L/K is purely inseparable, say [L : K ] = q = pa .
Step 2: By Noether Normalization, R is moduleﬁnite over a polynomial ring COMMUTATIVE ALGEBRA 239 k [t1 , . . . , td ]. If S is moduleﬁnite over k [t1 , . . . , td ], then certainly it is moduleﬁnite over the larger ring R. Thus we may assume without loss of generality that
R = k [t1 , . . . , td ], K = k (t1 , . . . , td ). In particular we may assume that R is integrally closed (in K ). For all a ∈ L, NL/K (a) = aq ∈ K . Let k ′ /k be the extension
obtained by adjoining the q th roots of the coeﬃcients of the minimal polynomials of
1/q
1/q
a ﬁnite set of generators of L/K , so k ′ /k is ﬁnite, so L ⊂ k ′ (t1 , . . . , td ). So it is
1/q
1/q
enough to show that the integral closure of k [t1 , . . . , td ] in k ′ (t1 , . . . , td ) is ﬁnite
over k [t1 , . . . , td ]. But in this case the integral closure can be computed exactly: it
1/q
1/q
is k ′ [t1 , . . . , td ] (indeed it is at least this large, and this ring is a UFD, hence
integrally closed), which is ﬁnite over k [t1 , . . . , td ].
18.3. The KrullAkizuki Theorem.
COMPLETE ME! ♣
19. Fractional ideals
19.1. Deﬁnition and ﬁrst properties.
Let R be an integral domain with fraction ﬁeld K . A fractional ideal of R is
a nonzero Rsubmodule I of K for which there exists 0 ̸= a ∈ R such that aI ⊂ R
1
– or equivalently, if I ⊂ a R.
When one is talking about fractional Rideals, one inevitably wants to compare
them to ideals of R in the usual sense, and for this it is convenient to speak of an
integral Rideal, i.e., an Rsubmodule of R.
Exercise: Show that any ﬁnitely generated Rsubmodule of K is a fractional ideal.
Comment: Some references deﬁne a fractional Rideal to be a ﬁnitely generated
Rsubmodule of K , but this seems wrong because we certainly want every nonzero
integral ideal of R to be a fractional ideal, but if R is not Noetherian then not every
integral ideal will be ﬁnitely generated. (It is not such a big deal because most of
these references are interested only in invertible fractional ideals – to be studied
shortly – and one of the ﬁrst things we will see is that an invertible fractional ideal
is necessarily ﬁnitely generated as an Rmodule.)
We denote the set of all fractional ideals of R by Frac(R). Let I, J ∈ Frac(R).
Theorem 418. Let I, J, K be fractional ideals in an integral domain R.
a) Then
I ∩ J = {x ∈ K  x ∈ I and x ∈ J },
I + J = {x + y  x ∈ I, y ∈ J,
n
∑
IJ = {
xi yi ,  xi ∈ I, yi ∈ J },
i=1 (I : J ) = {x ∈ K  xJ ⊂ I }
are all fractional ideals.
b) We may partially order Frac(R) under inclusion. Then the greatest lower bound
of I and J is I ∩ J and the least upper bound of I and J is I + J . 240 PETE L. CLARK c) If I ⊂ J , then IK ⊂ JK .
d) R itself is a fractional ideal, and R · I = R. Thus the fractional ideals form a
commutative monoid under multiplication.
Proof. a) It is immediately veriﬁed that I ∩ J , I + J , IJ and I : J are all Rsubmodules of K . It remains to be seen that they are nonzero and can be scaled
1
to lie inside R. Suppose I ⊂ a R and J ⊂ 1 R. Then:
b
1
0 I ⊂ I + J ⊂ ab R, so I + J is a fractional ideal.
1
0 I J ⊂ I ∩ J ⊂ ab R, so IJ and I ∩ J are fractional ideals.
Since I ∩ R is a fractional ideal, there exists a nonzero c ∈ R lying in I . Then for
1
y ∈ J , c y ∈ cR ⊂ I , so c ∈ (I : J ). Similarly, if 0 ̸= d ∈ J , then ad I : J ⊂ R. Parts
b
b
b), c) and d) can be easily veriﬁed by the reader.
Proposition 419. All the above operations on fractional ideals commute with localization: that is, if S ⊂ R• is a multiplicatively closed subset, then
(S −1 (I ∩ J ) = S −1 I ∩ S −I J,
S −1 (I + J ) = S −1 I + S −1 J,
S −1 (IJ ) = (S −1 I )(S −1 J ),
S −1 (I : J ) = (S −1 I : S −1 J ).
Exercise: Prove Proposition 419.
19.2. Principal Fractional Ideals.
A fractional ideal is said to be principal if it is of the form xR for some x = a
b ∈ K •. The following fact is easy but extremely important.
Proposition 420. For a fractional ideal I of R, TFAE:
(i) I is principal.
(ii) I is monogenic as an Rmodule.
(iii) I ∼R R.
=
Proof. By deﬁnition a monogenic Rmodule M is one of the form Rx for some
x ∈ M , so the equivalence of (i) and (ii) is immediate. Certainly R is monogenic
as an Rmodule. Conversely, suppose I = Rx for x ∈ K × . Then multiplication by
x−1 gives an isomorphism to R. (Another way to look at it is that a module M
over a domain R is isomorphic to R itself iﬀ it is monogenic and torsionfree, and a
principal fractional ideal has both of these properties.)
If xR is a principal fractional ideal, so is x−1 R, and we have
(xR)(x−1 R) = R.
Thus, in the monoid Frac(R), every principal fractional ideal xR is a unit, with
inverse x−1 R.
Let Prin(R) denote the set of all principal fractional ideals of the domain R.
Exercise: Show that Prin(R) is a subgroup of Frac(R), and we have a short exact sequence
1 → R× → K × → Prin(R) → 1. COMMUTATIVE ALGEBRA 241 Exercise: Deﬁne the ideal class monoid C (R) = Frac(R)/ Prin(R).
a) Show that C (R) is welldeﬁned as a commutative monoid.
b) Show that √(R) is trivial iﬀ R is a PID.
C
c) Let R = Z[ −3]. Show that C (R) is a ﬁnite commutative monoid which is not
a group.
It is somewhat disappointing that for a general domain, C (R) need only be a
commutative monoid. In the next section we “repair” this by deﬁning the Picard
group Pic(R).
19.3. Invertible Fractional Ideals and the Picard Group.
Like any monoid, Frac(R) has a group of units, i.e., the subset of invertible elements. Explicitly, a fractional ideal I is invertible if there exists another fractional
ideal J such that IJ = R. We denote the group Frac(R)× of invertible fractional
ideals by Inv(R).
Exercise: Let I1 , . . . , In be fractional ideals of R. Show that the product I1 · · · In
is invertible iﬀ each Ii is invertible. (Note: this has nothing to do with fractional
ideals, but is rather a fact about the units in a commutative monoid.)
It turns out that to every fractional ideal we can attach another fractional ideal I ∗
which will be the inverse of I iﬀ I is invertible. Namely, for I ∈ Frac(R), put
I ∗ = (R : I ) = {x ∈ K  xI ⊂ R}.
Immediately from the deﬁnition we get II ∗ ⊂ R.
Lemma 421. For a fractional ideal I , TFAE:
(i) I is invertible.
(ii) II ∗ = R.
Proof. (i) =⇒ (ii): As above, for any fractional ideal I we have II ∗ ⊂ R. Now
suppose there exists some fractional ideal J such that IJ = R, then
J ⊂ (R : I ) = I ∗ ,
so R = IJ ⊂ II ∗ . (ii) =⇒ (i) is obvious.
Proposition 422. Let I be an invertible fractional ideal. Then as an Rmodule I
is a rank one projective module.
Proof. Step 1: We show that an invertible fractional ideal I is a ﬁnitely generated
∑n
projective module. Since II ∗ = R, we may write 1 = i=1 xi yi with xi ∈ I and
yi ∈ I ∗ . For 1 ≤ i ≤ n, deﬁne fi ∈ Hom(I, R) be fi (x) = xyi . Then for all x ∈ I ,
∑
∑
x=
xxi yi =
xi fi (x).
i i By the Dual Basis Lemma, I is a projective Rmodule generated by x1 , . . . , xn .
Step 2: Recall that to show that I has rank one, we must show that for all p ∈
Spec R, Ip is free of rank one over Rp . But since projective implies locally free, we 242 PETE L. CLARK know that Ip is a free Rp module of some rank, and it is quite elementary to show
that for any ring R and any ideal I , I cannot be free of rank greater than one over
R. Indeed, if so I would have two Rlinearly independent elements x and y , which
is absurd, since yx + (−x)y = 0.
Conversely:
Proposition 423. Let I be a nonzero fractional ideal of R which is, as an Rmodule, projective. Then I is invertible.
Proof. We have the inclusion ι : II ∗ ⊂ R which we wish to show is an equality.
This can be checked locally: i.e., it is enough to show that for all p ∈ Spec R,
∗
ιp : Ip Ip → Rp is an isomorphism. By Proposition 419, it is equivalent to show
that Ip (Ip )∗ → Rp is an isomorphism, but since I is projective, by Kaplansky’s
Theorem Ip is free. As above, being a nonzero ideal, it is then necessarily free
of rank one, i.e., a principal fractional ideal xRp . It follows immediately that
(Ip )∗ = x−1 Rp and thus that the map is an isomorphism.
To sum up:
Theorem 424. Let I be a nonzero fractional ideal for a domain R. Then I is
invertible iﬀ it is projective, in which case it is necessarily projective of rank one.
For any Rmodule M , the Rdual is deﬁned to be M ∨ = Hom(M, R). There is a
canonical Rbilinear map T : M ∨ × M → R obtained by mapping (f, x) → f (x).
This induces an Rlinear map T : M ∨ ⊗R M → R. Let us say that an Rmodule
M is invertible if D is an isomorphism.
Proposition 425. Consider the following conditions on an Rmodule M .
(i) M is rank one projective.
(ii) M is invertible.
(iii) There exists an Rmodule N and an isomorphism T : M ⊗R N ∼ R.
=
Then (i) =⇒ (ii) =⇒ (iii) always, and (iii) =⇒ (i) if M is ﬁnitely generated.
Proof. (i) =⇒ (ii): As usual, we have a globally deﬁned map T : M ∨ ⊗ M → R
so that it suﬃces to check locally that is an isomorphism, but M is locally free so
this is easy.
(ii) =⇒ (iii) is immediate.
(iii) =⇒ (i): Since M is ﬁnitely generated, by Theorem 308 to show that M is
projective it is enough to show that for all p ∈ Spec R Mp is free of rank one. Thus
we may as well assume that (R, m) is a local ring with residue ﬁeld R/m = k . The
base change of the isomorphism T to R/m is an isomorphism (recall that tensor
product commutes with base change)
Tk : M/mM ⊗k N/mN → k.
This shows that dimk M/mM = dimk N/mN = 1, so in particular M/mM is
monogenic as an R/mmodule. By Nakayama’s Lemma the lift to R of any generator
x of M/mM is a generator of M , so M is a monogenic Rmodule and is thus
isomorphic to R/I for some ideal I . But indeed I = ann(M ) ⊂ ann(M ⊗R N ) =
ann(R) = 0, so M ∼ R/(0) = R is free of rank one.
=
Theorem 426. Let I and J be invertible fractional ideals. Then there is a canonical
isomorphism of Rmodules
∼
I ⊗R J → IJ. COMMUTATIVE ALGEBRA 243 Proof. The natural multiplication map I × J → IJ is Rbilinear so factors through
an Rmodule map m : I ⊗R J → IJ . Again, once we have a globally deﬁned map,
to see that it is an isomorphism it is enough to check it locally, i.e., that for all
p ∈ Spec R,
∼
mp : Ip ⊗Rp Jp → Ip Jp
and we are thus allowed to assume that I and J are principal fractional ideals. This
makes things very easy, and we leave the endgame to the reader.
Corollary 427. Let I and J be invertible fractional Rideals. TFAE:
(i) There exists x ∈ K × such that xI = J .
(ii) I ∼R J , i.e., I and J are congruent as Rmodules.
=
Proof. (i) =⇒ (ii): If J = xI , then multiplication by x gives an Rmodule
isomorphism from I to J .
(ii) =⇒ (i): Since I ∼R J we have
=
I −1 J ∼ I −1 ⊗R J ∼ I −1 ⊗R I ∼ II −1 = R.
=
=
=
By Proposition 420, I −1 J is a principal fractional ideal, i.e., there exists x ∈ K ×
such that I −1 J = xR. Multiplying through by I , we get xI = J .
Proposition 428. Let M be a rank one projective module over a domain R. Then
there exists a fractional Rideal I such that M ∼R I .
=
Proof. Let K be the fraction ﬁeld of R. Since M is projective, it is ﬂat, and so
tensoring the injection R → K with M we get an injection f : M = R ⊗R M →
M ⊗R K ∼ K , the last isomorphism since M is locally free of rank 1. Thus
=
∼
f : M → f (M ), and f (M ) is a ﬁnitely generated R submodule of K and thus a
fractional Rideal.
Putting together all the pieces we get the following important result.
Theorem 429. Let R be a domain. The following two commutative groups are
canonically isomorphic:
(i) Inv(R)/ Prin(R) with [I ][J ] := [IJ ].
(ii) Isomorphism classes of rank one projective Rmodules under tensor product.
We may therefore deﬁne the Picard group Pic R to be either the group of invertible
fractional ideals modulo principal fractional ideals under multiplication or the group
of isomorphism classes of rank one projective modules under tensor product.
19.4. More on invertible ideals.
Lemma 430. In any domain R, let P1 , . . . , Pk be a set of invertible prime ideals
and let Q1 , . . . , Ql be any set of prime ideals. Suppose that
k
∏
i=1 Pi = l
∏ Qj . j =1 Then i = j and there exists some permutation σ of the set {1, . . . , k } such that for
all 1 ≤ i ≤ k we have Pi = Qσ(i) .
In other words, prime factorization is unique for products of invertible primes. 244 PETE L. CLARK Proof. Assume without loss of generality that P1 does not strictly contain any
∏
Pi . Since j Qj ⊂ P1 , some Qj , say Q1 , is contained in P1 . Similarly, since
∏
i Pi ⊂ Q1 , there exists i such that Pi ⊂ Q1 . Thus Pi ⊂ Q1 ⊂ P1 . By our
assumption on the minimality of P1 , we have P1 = Pi = Q1 . We can thus cancel
P1 = Q1 and obtain the result by induction.
Lemma 431. Let R be an integrally closed Noetherian domain with fraction ﬁeld
K , and let I be a fractional Rideal. Then [I : I ] := {x ∈ F  xI ⊂ I } = R.
Proof. Clearly R ⊂ [I : I ]. Conversely, let x ∈ [I : I ]. Then I is a faithful
R[x]module which is ﬁnitely generated over R, so x is integral over R.
Lemma 432. Let R be a domain with fraction ﬁeld K , S ⊂ R \{0} a multiplicative
subset, and I, J fractional Rideals. a) We have (I + J )S = IS + JS .
b) (IJ )S = IS JS .
c) (I ∩ J )S = IS ∩ JS .
d) If I is ﬁnitely generated, then (I ∗ )S = (IS )∗ .
Proof. Parts a) and b) are immediate and are just recorded for future reference.
For part c), we evidently have (I ∩ J )S ⊂ IS ∩ JS . Conversely, let x ∈ IS ∩ JS ,
so x = si1 = sj2 with i ∈ I , j ∈ J , s1 , s2 ∈ S . Put b = a1 s2 = a2 s1 ∈ I ∩ J ; then
x = s1bs2 ∈ (I ∩ J )S , establishing part c). For part d), note ﬁrst that (I + J )∗ =
I ∗ ∩ J ∗ . Also if 0 ̸= x ∈ K , then (Rx)S = RS x. Hence if I = Rx1 + . . . + Rxn , then
∩n 1
∩n 1
IS = RS x1 + . . . + RS xn , so (IS )∗ = i=1 xi RS . On the other hand, I ∗ = i=1 xi R,
and thus part c) we have
n
∩1
∗
(I )S =
RS = (IS )∗ .
x
i=1 i
Lemma 433. Let R be a Noetherian domain of Krull dimension at most one. Let
I be a proper, nonzero ideal of R. Then (R : I ) R.
Proof. COMPLETE ME! ♣
The following result gives information about when a prime ideal is invertible.
Proposition 434. Let R be a Noetherian domain, and p a nonzero prime ideal of
R. If p is invertible, then it has height one and Rp is a DVR.
Proof. Since p is invertible, Rp is a Noetherian local domain with a principal maximal ideal pRp . By Theorem 412, Rp is a DVR, and thus p has height one.
19.5. Divisorial ideals. Let R be a domain with fraction ﬁeld K . Let Frac(R) be
the monoid of nonzero fractional Rideals in K , under multiplication.
For I, J ∈ Frac(R), we write I < J if every principal fractional ideal Ra which
is contained in I is also contained in J . This gives a preordering on Frac(R). Let
R be the associated equivalence relation, i.e., I ∼ J if I ≤ J and J ≤ I . We write
D(R) = Frac(R)/ ∼. Elements of D(R) are called divisors on R. For a fractional
ideal I , we denote its image in D(R) by div(I ), and for a principal fractional ideal
aR, we write simply div(a). Such elements are called principal divisors. COMMUTATIVE ALGEBRA 245 Let I ∈ Frac(R). By hypothesis there exists 0 ̸= d ∈ R such that I ⊂ d−1 R.
From this it follows that the intersection of all principal fractional ideals containing
I is itself a fractional ideal. We deﬁne this intersection to be
∩
˜
d−1 R.
I=
d∈R•  I ⊂d−1 R ˜
˜
˜
Evidently we have I = I , so I → I may be viewed as some sort of projection operator on fractional ideals. Indeed we say that a fractional ideal I is divisorial if
˜
˜
I = I . It is easy to see that I is indeed the least divisorial fractional ideal contain˜
ing I . Moreover, it is the unique divisorial fractional ideal such that div(I ) = div(I ).
Exercise: If I is a divisorial fractional ideal and x ∈ K • , then Ix is a divisorial
fractional ideal.
Exercise X.X: Let {Ii } be a family of divisorial fractional ideals such that I =
is nonzero. Then I is a divisorial fractional ideal. ∩ i Ii Lemma 435. Let I, J in Frac(A).
a) We have
∩
J :I=
x−1 J.
x∈I • b) If J is divisorial, then so is (J : I ).
Proof. a) Let y ∈ K . Then y ∈ J : I iﬀ yI ⊂ J iﬀ for all x ∈ I • , yx ∈ J iﬀ for all
x ∈ I • , y ∈ x−1 J .
b) Since J is divisorial, by Exercise X.X so is each x−1 J . Thus J : I is an intersection of divisorial fractional ideals which is, by Theorem 418, nonzero, hence by
Exercise X.X J : I is a divisorial fractional ideal.
Proposition 436. Let I, J ∈ Frac(R).
a) We have div(I ) = div(J ) iﬀ R : I = R : J .
˜
b) I = R : (R : I ).
Proof. a) Let P (I ) be the set of principal fractional ideals containing I . Thus
Rx ∈ P (I ) ⇐⇒ x−1 I ⊂ R ⇐⇒ x−1 ∈ R : I . Thus div(I ) = div(J ) ⇐⇒ P (I ) =
P (J ) ⇐⇒ R : I = R : J .
b) Since I (R : I ) ⊂ R, I ⊂ (R : (R : I )). Substituting (R : I ) for I in these relations,
we get
(R : I ) ⊂ R : (R : (R : I )).
On the other hand, from I ⊂ R : (R : I ) we get
(R : I ) ⊃ R : (R : (R : I )).
Thus
R : I = R : (R : (R : I ))),
thus using part a) we get div(I ) = div(R : (R : I )). However, by the previous
Lemma R : (R : I ) is divisorial, so it is the unique divisorial ideal whose divisor is
˜
equal to div(I ), i.e., I = R : (R : I ).
Exercise: Interpret the above results in terms of a Galois connections between
Frac(R) and D(R). 246 PETE L. CLARK Proposition 437. (Bourbaki, p. 477)
a) In D(R) every nonempty set which is bounded above admits a least upper bound.
Explicitly, if (Ii ) is a nonempty family of fractional ideals which is bounded above,
∩˜
then supi (div Ii ) = div( i Ii ).
b) In D(R) every nonempty set which is bounded below admits a greatest lower
bound. Explicitly, if (Ji ) is a nonempty family of fractional ideals which is bounded
∑
below, then inf i (div Ji ) = div( i Ii ).
c) D(R) is a lattice.
Proof. Bourbaki, p. 477. COMPLETE ME! ♣
19.6. The Divisor Class Group.
20. Dedekind domains
A Dedekind domain is an integral domain which is Noetherian, integrally closed,
and of dimension at most one. A Dedekind domain has dimension zero iﬀ it is a
ﬁeld. Although we endeavor for complete precision here (why not?), the reader
should be warned that in many treatments the zerodimensional case is ignored,
when convenient, in statements of results.
20.1. Characterization in terms of invertibility of ideals.
Theorem 438. For an integral domain R, TFAE:
(i) R is Dedekind: Noetherian, integrally closed of dimension at most one.
(ii) Every fractional Rideal is invertible.
Proof. (i) =⇒ (ii): Let R be a Noetherian, integrally closed domain of dimension
at most one, and let I be a fractional Rideal. Then II ∗ ⊂ R and hence also
II ∗ (II ∗ )∗ ⊂ R, so I ∗ (II )∗ ⊂ I ∗ . It follows from Lemma 431 that (II ∗ )∗ ⊂ R;
moreover, since II ∗ ⊂ R, Lemma 433 implies II ∗ = R, i.e., I is invertible.
(ii) =⇒ (i): Since invertible ideals are ﬁnitely generated, if every nonzero ideal
is invertible, then R is Noetherian. Let p be a nonzero, nonmaximal prime ideal
of R, so that there exists a maximal ideal m which 0
p
m. By the mantra
“to contain is to divide” for invertible fractional ideals, there exists some invertible
integral ideal I such that p = mI . Suppose that I ⊂ p. Then I = RI ⊃ mI = p, so
we would have p = I and then m = R, contradiction. Then there exists x ∈ m \ p
and y ∈ I \ p such that xy ∈ p, contradicting the primality of p.
b
Finally, we check that R is integrally closed: let x = c be a nonzero element of
the fraction ﬁeld K of R which is integral over R, so there exist a0 , . . . , an−1 ∈ R
such that
xn + an−1 xn−1 + . . . + a1 x + a0 = 0.
Let M be the Rsubmodule of K generated by 1, x, . . . , xn−1 ; since M is ﬁnitely
generated, it is a fractional Rideal. We have M 2 = M , and thus – since M is
invertible – M = R. It follows that x ∈ R.
Recall that a ring R is hereditary if every ideal of R is a projective Rmodule.
Corollary 439. A domain R is hereditary iﬀ it is a Dedekind domain.
Proof. By Theorem 438 a domain R is a Dedekind domain iﬀ every fractional ideal
of R is invertible, and clearly the latter condition holds iﬀ every nonzero integral
ideal of R is invertible. Moreover, by Theorem 424, a nonzero ideal of a ring is
invertible iﬀ it is projective as an Rmodule. COMMUTATIVE ALGEBRA 247 20.2. Ideal factorization in Dedekind domains.
Here we will show that in a Dedekind domain every nonzero integral ideal factors uniquely into a product of primes and derive consequences for the group of
invertible ideals and the Picard group. (The fact that factorization – unique or
otherwise! – into products of primes implies invertibility of all fractional ideals – is
more delicate and will be pursued later.)
Lemma 440. Let I be an ideal in a ring R. If there exist J1 , J2 ideals of R, each
strictly containing I , such that I = J1 J2 , then I is not prime.
Proof. Choose, for i = 1, 2, xi ∈ Ji \ I ; then x1 x2 ∈ I , so I is not prime.
Lemma 441. A nonzero ideal in a Noetherian domain contains a product of
nonzero prime ideals.
Proof. Assume not, and let I be a nonzero ideal which is maximal with respect
to the property of not containing a product of nonzero prime ideals. Then I is
not prime: there exist x1 , x2 ∈ R \ I such that x1 x2 ∈ I . Now put, for i = 1, 2,
Ii := ⟨I, xi ⟩, so that I
Ii and I ⊃ I1 I2 . By maximality of I , I1 ⊃ p1 · · · pr
and I2 ⊃ q1 · · · qs (with pi , qj prime for all i, j ), and then I ⊃ p1 · · · pr q1 ⊃ qs ,
contradiction.
Theorem 442. Every proper integral ideal in a Dedekind domain has a unique
factorization into a product of of prime ideals.
Proof. After Lemma 430 it suﬃces to show that a nonzero proper integral ideal I
in a Dedekind domain R factors into a product of primes. Suppose not, so the set
of ideals which do not so factor is nonempty, and (as usual!) let I be a maximal
element of this set. Then I is not prime, so in particular is not maximal: let p be
a maximal ideal strictly containing I , so I = pJ . Then J = p−1 I strictly contains
I so factors into a product of primes, hence I does.
Notation: If I is any nonzero integral ideal of I and p is any nonzero prime ideal
of a Dedekind domain R, then we may deﬁne nonnegative integers ordp (I ) via the
prime factorization
∏
I=
pordp (I ) .
p Note that the product extends formally over all primes, but as any given I is
divisible by only ﬁnitely many prime ideals, all but ﬁnitely many exponents are
zero, so it is really a ﬁnite product.
Corollary 443. Let R be a Dedekind domain.
a) The monoid M(R) of nonzero integral ideals is a free commutative monoid on
the set of nonzero prime ideals.
b) The fractional ideals form a free commutative group on the set of prime ideals:
⊕
Frac(R) =
Z.
0̸=p ∈ Spec R Proof. Part a) is simply the statement of unique factorization into prime elements in
any commutative monoid. In the group I (R) of all fractional ideals, the subgroup
G generated by the nonzero primes is a free commutative group on the primes:
this just asserts that for primes p1 , . . . , pr and integers n1 , . . . , nr , the equation 248 PETE L. CLARK pn1 · · · pnr = R implies n1 = . . . = nr = 0, which is easily seen – e.g. by localizing.
r
1
1
Since any fractional ideal J is of the form x I with I an integral ideal, decomposing
I and (x) into their prime factorizations expresses J as a Zlinear combination of
prime ideals, so Frac(R) = G.
Corollary 443 allows us to extend the deﬁnition of ordp to any fractional Rideal.
Since for a Dedekind domain there is no distinction between invertible fractional
ideals and all fractional ideals, the Picard group takes an especially simple form:
it is the quotient of the free abelian group Frac(R) of all fractional ideals modulo
the subgroup Prin(R) = K × /R× of principal fractional ideals. We therefore have
a short exact sequence
0 → Prin(R) → Frac(R) → Pic(R) → 0,
and also a slightly longer exact sequence
0 → R× → K × → Frac(R) → Pic(R) → 0.
Theorem 444. For a Dedekind domain R, TFAE:
(i) Pic(R) = 0.
(ii) R is a PID.
(iii) R is a UFD.
(iv) The set of nonprincipal prime ideals is ﬁnite.
Proof. Evidently each fractional ideal is principal iﬀ each integral ideal is principal:
(i) ≡ (ii). Since R has dimension at most one, (ii) ⇐⇒ (iii) by Proposition
380. Evidently (ii) =⇒ (iv), so the interesting implication is that (iv) implies the
other conditions. So assume that the set of (nonzero) nonprincipal prime ideals is
nonempty but ﬁnite, and enumerate them: p1 , . . . , pn . Let I be an integral ideal,
and suppose that
I = pa1 · · · pan qb1 · · · qbm .
m
n1
1
(As usual, we allow zero exponents.) By the Chinese Remainder Theorem we may
choose an α ∈ R such that ordpi (α) = ai for all i.60 Now consider the fractional
ideal (α−1 )I ; it factors as
(α−1 )I = qb1 · · · qbm rc1 · · · rcl ,
m1
1
l
where the ri ’s are some other prime ideals, i.e., disjoint from the pi ’s. But all of
the (fractional) ideals in the factorization of (α−1 )I are principal, so (α−1 )I = (β )
for some β ∈ K × and then I = (αβ ) is principal!
Remark: There exist onedimensional Noetherian domains with any ﬁnite number
√
of nonprincipal prime ideals. For instance, the ring√ [ −3] = Z[t]/(t2 + 3) has
Z
√
exactly one nonprincipal prime p2 := ⟨1 + −3, 1 − −3⟩.
60Note that we want equality, not just ord (α) ≥ a , so you should deﬁnitely think about
i
Pi
how to get this from CRT if you’ve never seen such an argument before. COMMUTATIVE ALGEBRA 249 20.3. Local characterization of Dedekind domains.
Theorem 445. Let R be an integral domain.
a) If R is Dedekind and S is a multiplicative subset, then S −1 R is Dedekind.
b) If R is a Dedekind domain and 0 ̸= p is a prime ideal of R, then Rp is a DVR.
Proof. The properties of being Noetherian, dimension at most one and integrally
closed are all preserved under localization, so part a) is immediate. Similarly, if
0 ̸= p is a prime ideal, then the localization Rp is a local, onedimensional integrally
closed Noetherian domain, hence by Corollary ?? a DVR, establishing b).
Exercise: Let R be a Dedekind domain with fraction ﬁeld K and p a nonzero prime
ideal of R.
a) Show that the map ordp : K × → Z deﬁned above is nothing else than the discrete valuation corresponding to the localization Rp .
b) Conversely, let v : K × → Z be a discrete valuation. Show that the valuation
ring Rv = v −1 (N) is the localization of R at some maximal ideal p.
Remark: Call a domain R for which Rm is a DVR for all maximal ideals m a
locally Dedekind domain.61
20.4. Factorization into primes implies Dedekind. We are now ready for our
last – and most diﬃcult – characterization theorem.
Theorem 446. Let R be a domain with the property that every nonzero proper
integral ideal is a product of prime ideals. Then R is Dedekind.
Proof. Step 1: Let p be an invertible prime of R. We show that p is maximal. Let
a ∈ R \ p, and suppose that ⟨a, p⟩ R. Let us then write
I1 := ⟨a, p⟩ = p1 · · · pm ,
I2 := ⟨a2 , p⟩ = q1 · · · qn ,
where the pi and qj are prime ideals. By assumption, I1 p, and, since p is prime,
we have also I2 p. Therefore each pi and qj strictly contains p. In the quotient
R = R/p we have
(a) = aR = p1 · · · pm
and
(a2 ) = a2 R = q1 · · · qn .
The principal ideals (a) and (a2 ) are invertible, and the pi and qj remain prime in
the quotient. Therefore, we have
q1 · · · qn = p2 · · · pm 2 .
1
Thus the multisets {{q1 , . . . , qn } and {p1 , p1 , . . . , pm , pm }} coincide, and pulling
back to R the same holds without the bars. Thus
2
I1 = ⟨a, p⟩2 = p2 · · · p2 = q1 · · · qn = ⟨a2 , p⟩,
1
m so
p ⊂ ⟨a, p⟩2 = a2 R + ap + p2 ⊂ aR + p2 .
61In the literature the term “almost Dedekind domain” is used, but we prefer to reserve this
for something else.) That nonNoetherian almost Dedekind domains exist is far from obvious;
such rings were ﬁrst constructed by N. Nakano [Nak53]. 250 PETE L. CLARK So if p ∈ p, p = ax + y with x ∈ R, y ∈ p2 , so ax ∈ p, and since a ∈ R \ p, x ∈ p.
Thus p ⊂ ap + p2 ⊂ p, so p = ap + p2 . Multiplication by p−1 gives R = a + p,
contrary to hypothesis. So p is maximal.
Step 2: Let p be any nonzero prime ideal in R, and 0 ̸= b ∈ p. Then p ⊃ bR and
bR = p1 · · · pm ,
with each pi invertible and prime. Thus by Step 1 the pi ’s are maximal. Since
p is prime we have p ⊃ pi for some i and then by maximality p = pi , hence p is
invertible. Since by assumption every proper integral ideal is a product of primes,
we conclude that every integral ideal is invertible, which, by Theorem 438 implies
that R is Dedekind.
Remark: Theorem 446 was ﬁrst proved by K. Matusita [Ma44]. It is often omitted
in introductory treatments because of its intricate proof, but it is a very enlightening result because it implies that the ideal theory in any nonDedekind ring will
have to be quite diﬀerent. For instance, any nonmaximal order in a number ﬁeld
will have “bad ideals” that do not factor into primes.
We summarize our work on characterizations of Dedekind domains:
Theorem 447. Let R be an integral domain which is not a ﬁeld. TFAE:
(i) R is Noetherian, integrally closed, and of Krull dimension one.
(ii) Every fractional (equivalently, every integral) Rideal is invertible.
(iii) R is Noetherian, and the localization at every maximal ideal is a DVR.
(iv) Every nonzero proper integral ideal factors into a product of prime ideals.
(iv′ ) Every nonzerp proper integral ideal factors uniquely into a product of primes.
A ring satisfying these equivalent conditions is called a Dedekind domain.
20.5. Generation of ideals in Dedekind domains.
Theorem 448. Let R be a Dedekind domain and I a nonzero ideal of R. Then
the quotient ring R/I is a principal Artinian ring.
∏r
Proof. Write I = i=1 pai . By the Chinese Remainder Theorem,
i
R/I ∼
= r
∏ R/pai .
i i=1 R/pai
i Each factor
is also a quotient of the localized ring Rp /pai , which shows that
i
it is Artinian and principal. Finally, a ﬁnite product of Artinian (resp. principal
ideal rings) remains Artinian (resp. a principal ideal ring).
This has the following striking consequence:
Theorem 449. (C.H. Sah) For an integral domain R, TFAE:
(i) R is a Dedekind domain.
(ii) For any nonzero ideal I of R and any nonzero element a ∈ I , there exists b ∈ I
such that I = ⟨a, b⟩.
Proof. The direction (i) =⇒ (ii) follows immediately from Theorem 448. Conversely, assume condition (ii) holds. By Theorem 447 it suﬃces to show that R
is Noetherian and that its localization at each nonzero prime ideal p is a DVR.
Certainly condition (ii) implies Noetherianity; moreover it continues to hold for
nonzero ideals in any localization. So let I be a nonzero ideal in the Noetherian COMMUTATIVE ALGEBRA 251 local domain (Rp , p). It follows that there exists b ∈ p such that p = I p + bRp . By
Nakayama’s Lemma, I = bRp , so Rp is a local PID, hence a DVR.
Proposition 450. ([J2, Ex. 10.2.11]) Let R be a Dedekind domain, I a fractional
ideal of R and J a nonzero integral ideal of R. Then there exists a ∈ I such that
aI −1 + J = R.
Proof. Let p1 , . . . , ps be the prime ideals of R dividing J . For each 1 ≤ i ≤ r, choose
ai ∈ I p1 · · · pr p−1 \ I p1 · · · pr . Put a = a1 + . . . + ar . We claim that aI −1 + J = R.
i
It is enough to check this locally. For every prime q ̸= pi , we have JRq = Rq . On
the other hand, for all 1 ≤ i ≤ r, aI −1 is not contained in pi , so its pushforward to
Rpi is all of Rpi .
20.6. Finitely generated modules over a Dedekind domain.
The aim of this section is to prove the following important result.
Theorem 451. Let M be a ﬁnitely generated module over a Dedekind domain R.
Then:
a) P = M/M [tors] is a ﬁnitely generated projective Rmodule, say of rank r.
b) If r = 0 then of course M = M [tors]. If r ≥ 1 then
M ∼ M [tors] ⊕ P ∼ M [tors] ⊕ Rr−1 ⊕ I,
=
=
with I a nonzero ideal of R.
c) The class [I ] of I in Pic R is an invariant of M .
d) There exists N ∈ Z+ , maximal ideals pi and positive integers ni such that
M [tors] ∼
= N
⊕ R/pni .
i i=1 Much of the content of the main theorem of this section lies in the following converse
of Proposition 16b) for ﬁnitely generated modules over a Dedekind domain.
Theorem 452. For a ﬁnitely generated module M over a Dedekind domain, TFAE:
(i) M is projective.
(ii) M is ﬂat.
(iii) M is torsionfree.
Proof. Of course (i) =⇒ (ii) =⇒ (iii) for modules over any domain, and we have
seen that (i) ≡ (ii) for ﬁnitely generated modules over a Noetherian ring. So it
suﬃces to show (iii) =⇒ (i).
Suppose R is a Dedekind domain and M is a ﬁnitely generated nonzero torsionfree Rmodule. By Proposition 16c), we may assume that M ⊂ Rn for some n ≥ 1.
We prove the result by induction on n. If n = 1, then M is nothing else than a
nonzero ideal of R, hence an invertible module by Theorem XX and thus a rank one
projective module by Theorem XX. So we may assume that n > 1 and that every
ﬁnitely generated torsionfree submodule of Rn−1 is projective. Let Rn−1 ⊂ Rn be
the span of the ﬁrst n − 1 standard basis elements. Let πn : Rn → R be projection
onto the nth factor, and consider the restriction of πn to M :
n
0 → M ∩ Rn−1 → M → πn (M ) → 0. π 252 PETE L. CLARK Put I = πn (M ). Then I is an ideal of R, hence projective, so the sequence splits:
M → (M ∩ Rn−1 ) ⊕ I.
Now M ∩ Rn−1 is a torsionfree, ﬁnitely generated (since M is ﬁnitely generated and
R is Noetherian) submodule of Rn−1 , hence is projective by induction. Certainly
a direct sum of projective modules is projective, so we’re done.
The method of proof immediately yields the following important corollary:
Corollary 453. Let P be a ﬁnitely generated rank r projective module over a
⊕r
Dedekind domain R. Then we have a direct sum decomposition P ∼
=
i=1 Ii ,
where each Ii is a nonzero rank one projective module (equivalently, isomorphic to
a nonzero ideal of R).
Now let M a ﬁnitely generated module over the Dedekind domain R. We have the
exact sequence
0 → M [tors] → M → M/M [tors] → 0.
Put P := M/M [tors]. Then P is ﬁnitely generated (clearly) and torsionfree (by
Proposition 16a), hence it is a ﬁnite rank projective module (by Theorem 452) and
the sequence splits:
M ∼ M [tors] ⊕ P.
=
Lemma 454. I1 , . . . , In be fractional ideals in the Dedekind domain R. Then the
⊕n
Rmodules i=1 Ii and Rn−1 ⊕ I1 · · · In are isomorphic.
Proof. We will prove the result when n = 2. The general case follows by an easy
induction argument left to the reader.
−
Choose 0 ̸= a1 ∈ I1 . Applying Proposition 450 with I = I2 and J = a1 I1 1 ⊂ R,
−
−
−
that there exists a2 ∈ I2 such that a1 I1 1 + a2 I2 1 = R. That is there exist bi ∈ Ii 1
such that a1 b1 + a2 b2 = 1. The matrix
[
b1 −a2
b2 a1
is invertible with inverse
A−1 = [ a1
−b2 a2
b1 . For (x1 , x2 ) ∈ I1 I2 , we have
y1 = x1 b1 + x2 ∈ R, y2 = −x1 a2 + x2 a1 ∈ I1 I2 .
On the other hand, if y1 ∈ R and y2 = c1 c2 ∈ I1 I2 , then
x1 = a1 y1 − b2 c1 c2 ∈ I1 , x2 = a2 y1 + b1 c1 c2 ∈ I2 .
Thus [x1 x2 ] → [x1 x2 ]A gives an Rmodule isomorphism from I1 ⊕ I2 to R ⊕ I1 I2 .
Thus we may write
M = M [tors] ⊕ M/M [tors] ∼ M [tors] ⊕
= r
⊕ Ii = M [tors] ⊕ Rr−1 ⊕ (I1 · · · Ir ), i=1 which establishes Theorem 451a).
As for part b) of the theorem, let T be a ﬁnitely generated torsion Rmodule. Note
that the statement of the classiﬁcation is identical to that of ﬁnitely generated torsion modules over a PID. This is no accident, as we can easily reduce to the case of a COMMUTATIVE ALGEBRA 253 PID – and indeed to that of a DVR, which we have already proven (Theorem 414).
Namely, let I be the annihilator of T , and (assuming T ̸=∏ as we certainly may)
0,
∏r
⊕r
r
write I = i=1 pai . Then T is a module over R/I ∼ R/ ⊕ pri ∼ i=1 R/pai .
=
i
i
i=1 i =
r
By Exercise X.X in §3.1, T naturally decomposes as T = i=1 Ti , where Ti is a
module over R/pai . This gives the primary decomposition of T . Moreover, each Ti
i
is a module over the DVR Rp , so Theorem 414 applies.
Corollary 455. For any Dedekind domain R, the Picard group Pic R is canonically
isomorphic to the reduced K0 group K0 (R).
Proof. Let P be a ﬁnitely generated projective Rmodule of rank r ≥ 1. According
to Theorem 451c) the monoid of isomorphism classes of ﬁnitely generated projective
Rmodules is cancellative : this means that the canonical map φ : Pic(R) → K0 (R)
φ is injective. It follows easily that the composite map Φ : Pic(R) → K0 (R) → K0 (R)
is an injection: indeed, for φ(I ) to be killed in K0 (R) but not K0 (R) it would have
to be a fractional ideal which has rank zero as an Rmodule, and there are no
such things. Now an arbitrary nonzero ﬁnitely generated projective Rmodule is
isomorphic to Rr−1 ⊕ I , hence becomes equal to the class of the rank one module I
in K0 (R), so Φ is surjective. To check that it is a homomorphism of groups we may
look on a set of generators – namely, the classes of rank one projective modules.
Let us use [P ] for the class of the projective module P in K0 (R) and [[P ]] for its
image in K0 (R). Then by Lemma 454 we have
Φ([I1 ⊗ I2 ]) = [[I1 ⊗ I2 ]] = [[I1 I2 ]] = [[R ⊕ I1 I2 ]] = [[I1 ⊕ I2 ]] = [[I1 ]] + [[I2 ]]. ¨
21. Prufer domains
A Pr¨ fer domain is an integral domain in which each nonzero ﬁnitely generated
u
ideal is invertible. Since principal ideals are invertible, any B´zout domain is a
e
Pr¨fer domain.
u
21.1. Characterizations of Pr¨ fer Domains.
u
One might be forgiven for thinking the invertibility of ﬁnitely generated ideals
is a somewhat abstruse condition on a domain. The following result shows that, on
the contrary, this determines a very natural class of domains.
Theorem 456. (Characterization of Pr¨fer Domains) For an integral domain R,
u
TFAE:
(i) R is a Pr¨fer domain.
u
(ii) Every nonzero ideal of R generated by two elements is invertible.
(iii) Finitely generated ideals are cancellable: if A, B, C are ideals of R and A is
ﬁnitely generated, then AB = AC =⇒ B = C .
(iv) For every nonzero prime ideal p of R, Rp is a valuation ring.
(v) For all ideals A, B, C of R, A(B ∩ C ) = AB ∩ AC .
(vi) For all ideals A, B of R, (A + B )(A ∩ B ) = AB .
(vii) If A and C are ideals of R with C ﬁnitely generated and A ⊂ C , then there
exists an ideal B of R such that A = BC . 254 PETE L. CLARK (vii) For all ideals A, B, C of R with C ﬁnitely generated, we have
(A + B : C ) = (A : C ) + (B : C ).
(viii) For all ideals A, B, C of R with C ﬁnitely generated, we have
(C : A ∩ B ) = (C : A) + (C : B ).
(ix) For all ideals A, B, C of R, A ∩ (B + C ) = A ∩ B + A ∩ C .
Proof. [LM, Thm. 6.6].
Theorem 457. Let R be a domain.
a) Suppose R is a GCDdomain. Then R is a Pr¨fer domain iﬀ it is a B´zout
u
e
domain.
b) If R is a Pr¨fer domain and a UFD, then it is a PID.
u
Proof. a) Since principal ideals are invertible, any B´zout domain is a Pr¨fer doe
u
main. Conversely, suppose R is a GCDdomain and a Pr¨fer domain. Let x, y ∈ R•
u
and let d be a GCD of x, y . Certainly we have (d) ⊃ ⟨x, y ⟩. Thus ι : ⟨x, y ⟩ → (d)
is a homomorphism of Rmodules which we want to show is an isomorphism. By
the LocalGlobal Principle for Module Homomorphisms it is enough to show that
for all p ∈ Spec R, ιp is an isomorphism of Rp modules, i.e., ⟨x, y ⟩Rp = ⟨d⟩Rp . By
Proposition 356, d is again the GCD of x and y in the valuation ring Rp (equivalently, the valuation of d is the minimum of the valuations of x and y ) so that the
principal ideal ⟨x, y ⟩Rp is generated by ⟨d⟩Rp .
b) Suppose R is a Pr¨fer UFD. By part a) it is also a B´zout domain and by XXX
u
e
a B´zout UFD is a PID.
e
Proposition 458. For a Pr¨fer domain R, TFAE:
u
(i) R is a B´zout domain.
e
(ii) Pic(R) = 0.
Proof. In the Pr¨fer domain R, an ideal I is invertible iﬀ it is ﬁnitely generated.
u
So (i) and (ii) each assert that every ﬁnitely generated ideal is principal.
Proposition 459. A Pr¨fer domain is integrally closed.
u
Proof. In Theorem 438 we showed that a domain in which all fractional Rideals are
invertible is integrally closed. In the proof we only used the invertbility of ﬁnitely
generated fractional ideals, so the argument works in any Pr¨fer domain.
u
Exercise: Give a proof of Corollary 459 using the local nature of integral closure.
21.1.1. A Chinese Remainder Theorem for Pr¨fer domains.
u
Recall that we have a Chinese Remainder Theorem which is valid in any ring:
Theorem 126. There is however another, “more elementwise” version of the Chinese Remainder Theorem which holds in a domain R iﬀ R is a Pr¨fer domain.
u
Let R be a ring, let I1 , . . . , In be a ﬁnite sequence of ideals in R and let x1 , . . . , xn
be a ﬁnite sequence of elements in R. We may ask: when is there an element x ∈ R
such that x ≡ xi (mod Ii ) for all i?
If we assume that the ideals Ii are pairwise comaximal, then this holds in any
ring by the Chinese Remainder Theorem (Theorem 126). But suppose we drop COMMUTATIVE ALGEBRA 255 that condition. Then, if such an x exists, we have x − xi ∈ Ii for all i, hence for all
i and j ,
(25) xi − xj = (x − xj ) − (x − xi ) ∈ Ii + Ij . Thus we get a necessary condition (which is vacuous when the Ii ’s are pairwise
comaximal). Let us say that a ring has property ECRT (elementwise Chinese Remainder Theorem) if given any ideals I1 , . . . , In and elements x1 , . . . , xn satisfying
(25), there exists x ∈ R such that x ≡ xi (mod Ii ) for all i.
Exercise: Show that a PID satisﬁes property ECRT.
Theorem 460. For a domain R, TFAE:
(i) R satisﬁes property ECRT.
(ii) R is a Pr¨fer domain.
u
Proof. . . .
21.2. Modules over a Pr¨ fer domain.
u
Proposition 461. A domain R is a semihereditary iﬀ it is a Pr¨fer domain.
u
Exercise: Prove Proposition 461.
Lemma 462. Let R be a domain, and let M be a ﬁnitely generated torsionfree
Rmodule. Then M is a submodule of a ﬁnitely generated free module.
Proof. Since M is torsionfree, the natural map M → M ⊗R K is injective, and
ι : M ⊗R K ∼ K n for some n ∈ N. Since M is ﬁnitely generated, there exists
=
x ∈ R• such that the image of xM in M ⊗R K is contained in Rn , and thus
ι ◦ (x•) : M → Rn .
Theorem 463. For an integral domain R, TFAE:
(i) Every torsionfree Rmodule is ﬂat.
(ii) Every ﬁnitely generated torsionfree Rmodule is projective.
(iii) R is a Pr¨fer domain.
u
Proof. (i) =⇒ (ii): In particular every ﬁnitely generated torsionfree Rmodule is
ﬂat. Since R is a domain, by Corollary 309 every ﬁnitely generated ﬂat module is
projective.
(ii) =⇒ (iii): An ideal of a domain is a torsionfree Rmodule, so if (ii) holds every
ﬁnitely generated ideal of R is projective, hence invertible.
(iii) =⇒ (i): Let R be a Pr¨ fer domain and M a torsionfree Rmodule. Then
u
M = limi Mi is the direct limit of its ﬁnitely generated submodules, hence a direct
−
→
limit of ﬁnitely generated torsionfree modules Mi . By Lemma 462, each Mi is a
ﬁnitely generated submodule of a free Rmodule. By Theorems 461 and 70, each
Mi is projective, hence ﬂat. Thus M is a direct limit of ﬂat modules, hence is itself
a ﬂat module by Corollary 90.
22. Structure of overrings 256 PETE L. CLARK 22.1. Introducing overrings.
Let R be an integral domain with fraction ﬁeld K . By an overring of R we
mean a subring of K containing R, i.e., a ring T with R ⊂ T ⊂ K . (We allow
equality.) This is standard terminology among commutative algebraists, but we
warn that someone who has not heard it before will probably guess incorrectly at
its meaning: one might well think that “T is an overring of R” would simply mean
that “R is a subring of T ”.
The basic idea of the study of overrings is the following: if R is a very simple
ring, then (i) each of its overrings will have very nice properties and (ii) the set of
all overrings of R will have simple, transparent structure.
Example: Suppose R is a DVR. Then R
erly in between. K and there are no overrings prop Example: Suppose R is a PID. Then the overrings of R correspond to the subsets Σ ⊂ MaxSpec R of nonzero prime ideals: every overring is of the form RΣ , in
which we invert one (equivalently, all) generators of all nonzero prime ideals p = (p)
∼
lying outside of Σ. Thus we get a canonical isomorphism MaxSpec RΣ → Σ.
Note that for every multiplicative subset S of R, the localization S −1 R is an overring of R. As we have just seen, for PIDs the converse is true. So we may ask:
Question 5. For which domains R is every overring a localization?
22.2. Overrings of Dedekind domains.
Let R be a Dedekind domain with fraction ﬁeld K : we assume R ̸= K . Let
ΣR denote the set of height one primes of R. For W ⊂ ΣR , we deﬁne
∩
RW =
Rp
p∈W and also
RW = ∩ Rp . p∈ΣR \W When W = {p} consists of a single element, we we write Rp for R{p} .
Lemma 464. a) For all p ∈ ΣR , there exists fp ∈ Rp \ R.
b) The mapping W → RW is injective.
Proof. a) Choose x ∈ K \ Rp , and let S be the ﬁnite set of maximal ideals q
distinct from p such that ordq (x) < ∏ For each q ∈ S , let yq ∈ q \ p. Let
0.
N = maxq∈S − ordq (x), and put fp = ( q∈S yq )N x.
b) Suppose W1 and W2 are distinct subsets of ΣR . After relabelling if necessary,
we may assume that there exists p ∈ W2 \ W1 . By part a), there exists fp ∈ Rp \ R
and thus fp ∈ RW1 \ RW2 .
Proposition 465. Let R be a Dedekind domain with fraction ﬁeld K , and let T
be an overring of R. Write ι : R → T for the inclusion map.
a) For every P ∈ ΣT , T = RP∩R . COMMUTATIVE ALGEBRA b)
c)
d)
e) T is itself a Dedekind domain.
ι∗ : ΣT → ΣR is an injection.
For all P ∈ ΣT , ι∗ ι∗ P = P .
ι∗ identiﬁes ΣT with the subset of p ∈ ΣR such that pT 257 T. Proof. a) Put p = P ∩ R. There exist x, y ∈ R• such that ∈ P . Then 0 ̸= x =
y ( x ) ∈ p, so p is a nonzero prime ideal of R. Thus TP contains the DVR Rp and is
y
properly contained in its fraction ﬁeld, so TP = Rp .
b) By the KrullAkizuki Theorem, T is a Noetherian domain of Krull dimension
one. By part a), the localization of T at every prime is a DVR. So T is integrally
closed and is thus a Dedekind domain.
c) For distinct P1 , P2 ∈ ΣT , the localizations TP1 , TP2 are distinct DVRs. But by
part a), putting pi = ι∗ (Pi ), we have Rpi = TPi for i = 1, 2, so p1 ̸= p2 .
d) Suppose p = ι∗ (P ). Then
x
y (ι∗ ι ∗ P )TP = pTP = P TP .
∗ By part c), ι ∗ ι P is not divisible by any prime other than P , so ι∗ p = P .
e) This follows immediately and is stated separately for later use.
Lemma 466. Let R be an integrally closed domain with fraction ﬁeld K , and let
T be an overring of R.
a) The relative unit group T × /R× is torsionfree.
b) Suppose that R is a Dedekind domain, p ∈ ΣR and T = Rp . TFAE:
(i) T × /R× ∼ Z.
=
(ii) T × R× .
(iii) [p] ∈ Pic(R)[tors].
(iv) There exists x ∈ R which is contained in p and in no maximal ideal q ̸= p.
Proof. a) Since R is integrally closed, all ﬁnite order elements of K × (i.e., roots of
unity in K ) lie in R and a fortiori in T : R× [tors] = T × [tors]. On the other hand,
let x ∈ T × be of inﬁnite order such that xn ∈ R× for some n ∈ Z+ . Again integral
closure of R implies x ∈ R, and then xn ∈ R× =⇒ x ∈ R× .
b) (i) =⇒ (ii) is clear.
(ii) =⇒ (iii): Let x ∈ K × . Then x ∈ T × iﬀ Rx = pa for some a ∈ Z, and x ∈ R×
iﬀ a = 0. Therefore (ii) holds iﬀ some power of p is principal, which is to say that
the class of p ∈ Pic R is torsion.
(iii) =⇒ (i): Let a be the least positive integer such that pa is principal. Thus
pa = xR with x uniquely determined modulo R× . It follows that T × is generated
by R× and x, so T × /R× is a nontrivial cyclic group. By part a) it is also torsionfree
so T × /R× ∼ Z.
=
(iii) =⇒ (iv): If pa = xR, then x lies in p but in no other maximal ideal q.
(iv) =⇒ (iii): If a = vp (x), then a > 0 and xR = pa .
Remark: Part (iv) of Lemma 466 was added following an observation of H. Knaf.
Theorem 467. (Goldman [Gol64]) For a Dedekind domain R, TFAE:
(i) Pic R is a torsion group.
(ii) Every overring of R is a localization.
Proof. (i) =⇒ (ii): Let p ∈ ΣR . By Lemma 464 Rp is a proper overring of R, so
by assumption Rp is a localization of R and thus has a strictly larger unit group.
By Lemma 466 this implies that [p] ∈ Pic(R)[tors]. Since Pic(R) is generated by 258 PETE L. CLARK the classes of the nonzero prime ideals, it follows that Pic R is torsion.
(ii) =⇒ (i): Let T be an overring of R, and put S = R ∩ T × . We want to show that
T = S −1 R. That S −1 R ⊂ T is clear. Conversely, let x ∈ T , and write xR = ab−1
with a, b coprime integral ideals of R: a + b = R. Thus aT + bT = T whereas
aT = xbT ⊂ bT , so bT = T and hence also bn T = T for all n ∈ Z+ . Since Pic R is
torsion, there exists n ∈ Z+ with bn = bR. It follows that bT = T and thus b ∈ S .
Now xb = a ⊂ R, so xb ∈ R. Thus x ∈ S −1 R, and we conclude T ⊂ S −1 R.
The last result that we want to prove about overrings of Dedekind domains is a
complete classiﬁcation: let T be an overring of the Dedekind domain R. As we have
seen, pulling back via the inclusion ι allows us to view the set of maximal ideals
of T as a subset, say W of the set ΣR of maximal ideals of R. Moreover, for each
W ⊂ ΣR there is at least one overring T with ΣT = W , namely RW . We claim
that in fact these are all the overrings of R, or more succinctly:
Theorem 468. For any overrring T of the Dedekind domain R, we have
T = Rι∗ (ΣT ) .
The proof (that I know) of this theorem requires the study overrings of more general
P¨ufer domains, to which we turn in the next section.
r
22.3. Overrings of Pr¨ fer Domains.
u
Theorem 469. Let R be an integral domain with fraction ﬁeld K , and let T be an
overring of R. TFAE:
(i) For every prime ideal p of R, either pT = T or T ⊂ Rp .
(ii) For every x, y ∈ K × with x ∈ T , ((y ) : (x))T = T .
y
(iii) T is a ﬂat Ralgebra.
Proof. . . .
Proposition 470. Let R be an integral domain with fraction ﬁeld K and consider
rings R ⊂ T ⊂ T ′ ⊂ K .
a) If T ′ is ﬂat over R, then T ′ is ﬂat over T .
b) If T ′ is ﬂat over T and T is ﬂat over R, then T ′ is ﬂat over R.
c
Proof. a) Suppose T ′ is ﬂat over R. Let a, b ∈ T be such that a ∈ T ′ . Write a = s ,
b
c
b = d with c, d, s ∈ R. Then d ∈ T ′ , so by Theorem 469, ((d) : (c))T ′ = T ′ . Hence
s
1 = t1 u1 + . . . + tk uk for some ti ∈ T ′ and ui ∈ R with ui c ∈ (d) for all i. Then
there exists zi ∈ R such that ui c = dzi , so ui a = zi d = zi b ∈ T b for all i. So
s
(T b : T a)T ′ = T ′ . Applying Theorem 469 again, we conclude that T ′ is ﬂat over T .
b) This holds for any tower of ring extensions R1 ⊂ R2 ⊂ R3 , since M ⊗R1 R3 ∼
=
(M ⊗R1 R2 ) ⊗R2 R3 . Proposition 471. For an overring T of an integral domain R, TFAE:
(i) T is ﬂat over R.
(ii) For all p ∈ MaxSpec T , Tp = Rp∩R .
∩
(iii) T = p∈MaxSpec T Rp .
Proof. . . .
Proposition 472. Let T be an overring of a domain R which is both integral and
ﬂat over R. Then R = T . COMMUTATIVE ALGEBRA 259 Proof. Let x, y ∈ R be such that x ∈ T . Then by Theorem 469, ((y ) : (x))T = T .
y
Let p ∈ MaxSpec R. By Theorem 322, there exists a prime (in fact maximal by
Corollary 325, but this is not needed here) ideal P of T lying over p. Since pT ⊂ P ,
we have pT T . Therefore ((y ) : (x)) is not contained in any maximal ideal of R,
so ((y ) : (x)) = R. It follows that x ∈ (y ), i.e., x = ay for some a ∈ R, so that
x
y ∈ R. Thus R = T .
Theorem 473. For an integral domain R, TFAE:
(i) R is a Pr¨fer domain.
u
(ii) Every overring of R is a ﬂat Rmodule.
Proof. (i) =⇒ (ii): An overring is a torsionfree Rmodule, so this follows immediately from Theorem 463.
(ii) =⇒ (i): Let p be a maximal ideal of R. By Proposition 470, every overring of
Rp is ﬂat. Let a, b ∈ Rp , and suppose that aRp ̸⊂ bRp . Then (bRp : aRp ) ̸= Rp ;
since Rp is local, this implies (bRp : aRp ) ⊂ pRp . Now consider the ring Rp [ a ].
b
This is an overring of Rp , hence ﬂat. Since a ∈ Rp [ a ] we have by Theorem 469
b
b
a
a
(bRp : aRp )Rp [ ] = Rp [ ].
b
b
Thus there exist elements x1 , . . . , xn ∈ (bRp : aRp and b1 , . . . , bn ∈ Rp [ a ] such that
b
x1 b1 + . . . + xn bn = 1. . . .
Corollary 474. Every overring of a Pr¨fer domain is a Pr¨fer domain.
u
u
Proof. Let R be a Pr¨fer domain and T be an overring of R. Then every overring
u
T ′ of T is in particular an overring of R, so T ′ is ﬂat over R. By Proposition ??a),
T ′ is also ﬂat over T . Therefore every overring of T is ﬂat over T , so by Theorem
473, T is a Pr¨fer domain.
u
Finally we give a result which generalizes the (as yet unproven) Theorem 468.
Namely, for R a domain, let W be a subset of MaxSpec R and put
∩
RW =
Rp .
p∈W Theorem 475. Let R be a Pr¨fer domain, T an overring of R, and put
u
W = {p ∈ MaxSpec(R)  pT T }. Then T = RW .
Proof. . . .
22.4. Kaplansky’s Theorem (III).
Theorem 476. (Kaplansky [K]) Let R be a Dedekind domain with fraction ﬁeld
K , and let K be an algebraic closure of K . Suppose that for every ﬁnite extension
L/K , the Picard group of the integral closure RL of R in L is a torsion abelian
e
group. Then the integral closure S of R in K is a B´zout domain.
Proof. Let I = ⟨a1 , . . . , an ⟩ be a ﬁnitely generated ideal of S . Then L = K [a1 , . . . , an ]
is a ﬁnite extension of K . Let RL be the integral closure of R in L, and let
IL = ⟨a1 , . . . , an ⟩RL . By hypothesis, there exists k ∈ Z+ and b ∈ RL such that
k
IL = bRL . Let c be a k th root of b in S and let M = L[c]. Thus in the Dedekind
domain RM we have (IL RM )k = (ck ), and from unique factorization of ideals we
deduce IL RM = cRM . Thus I = IL RM S = cRM S = cS is principal. 260 PETE L. CLARK Recall the basic fact of algebraic number theory that for any number ﬁeld K , the
Picard group of ZK is ﬁnite. This shows that the ring R = Z satisﬁes the hypotheses of Theorem 476. We deduce that the ring of all algebraic integers Z is a B´zout
e
domain: Theorem 129.
Exercise X.X: State a function ﬁeld analogue of Theorem 129 and deduce it as
a special case of Theorem 476.
22.5. Every commutative group is a Picard group.
23. Krull domains
Let R be an integral domain with fraction ﬁeld K . Let Σ(R) be the set of height
one prime ideals of R. We say that R is a Krull domain if it satisﬁes the following
properties:
(KD1) For each p ∈ Σ(R), Rp is a DVR.
∩
(KD2) R = p∈Σ(R) Rp .
(KD3) For 0 ̸= x ∈ R, {p ∈ Σ(R)  f ∈ p} is ﬁnite.
In general, let K be a ﬁeld, and let {Ri } be a family of subrings of K , and put
∩
R = i Ri . We say that the family has ﬁnite character (FC) if for any 0 ̸= x ∈ R,
×
x ∈ Ri for all but ﬁnitely many i ∈ I .
Exercise X.X: If {Ri }i∈I is a ﬁnite character family of subrings of a ﬁeld K and
J ⊂ I , show that the subfamily {Ri }i∈J also has ﬁnite character.
Theorem 477. For a domain R, TFAE:
(i) R is a Krull domain.
(ii) There exists a ﬁeld K containing R, and a ﬁnite character family of DVR’s
∩
{Vi } inside K such that R = i Vi .
Proposition 478. a) A Krull domain is normal.
b) Any normal Noetherian domain is a Krull domain.
Proof: a) By (KD1) and (KD2), R is the intersection inside its ﬁeld of fractions of
a family of normal domains, hence it is itself normal. b) . . .
Proposition 479. Let R be a Krull domain with fraction ﬁeld K , and let k be a
subﬁeld of K . Then R ∩ k is a Krull domain.
Proof: . . .
Proposition 480. Let R be a Krull domain with fraction ﬁeld K , and let L/K be
a ﬁnite extension. Then the normalization of R in L is a Krull domain.
Proof: . . .
Proposition 481. Let {Ri }i∈I be a family of Krull subdomains of the ﬁeld K . If
∩
the family satisﬁes the ﬁnite character property, then R = i Ri is a Krull domain.
If R is a Krull domain, with set of height ∩ primes Σ(R), and T ⊂ Σ(R) is an
one
arbitrary subset, consider the ring RT := p∈T Rp . Evidently RT is an overring
of R, i.e., is intermediate between R and its ﬁeld of fractions. By Exercise X.X, COMMUTATIVE ALGEBRA 261 the family {Rp }p∈T also has ﬁnite character, so by Theorem 477 RT is also a Krull
domain. An overring of a Krull domain arising from a subfamily T in this way is
called a subintersection.
In a Krull domain, localizations are subintersections:
Proposition 482. Let R be a Krull domain, and S a multiplicatively closed subset.
Let T ⊂ ∩ R) be the set of height one primes p which are disjoint from S ; then
Σ(
S −1 R = p∈T Rp . In particular, a localization of a Krull domain is a Krull domain.
Theorem 483. A polynomial ring in any (ﬁnite or inﬁnite) number of indeterminates over a Krull domain is a Krull domain.
In particular, a polynomial ring in inﬁnitely many indeterminates over a ﬁeld is a
nonNoetherian Krull domain.
24. Connections with model theory
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